Sum of squared logarithms - an inequality relating positive definite matrices and their matrix logarithm

  • Mircea Bîrsan1, 2,

    Affiliated with

    • Patrizio Neff1Email author and

      Affiliated with

      • Johannes Lankeit1

        Affiliated with

        Journal of Inequalities and Applications20132013:168

        DOI: 10.1186/1029-242X-2013-168

        Received: 21 January 2013

        Accepted: 28 March 2013

        Published: 12 April 2013

        Abstract

        Let y 1 , y 2 , y 3 , a 1 , a 2 , a 3 ( 0 , ) http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq1_HTML.gif be such that y 1 y 2 y 3 = a 1 a 2 a 3 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq2_HTML.gif and

        y 1 + y 2 + y 3 a 1 + a 2 + a 3 , y 1 y 2 + y 2 y 3 + y 1 y 3 a 1 a 2 + a 2 a 3 + a 1 a 3 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equa_HTML.gif

        Then

        ( log y 1 ) 2 + ( log y 2 ) 2 + ( log y 3 ) 2 ( log a 1 ) 2 + ( log a 2 ) 2 + ( log a 3 ) 2 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equb_HTML.gif

        This can also be stated in terms of real positive definite 3 × 3 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq3_HTML.gif-matrices P 1 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq4_HTML.gif, P 2 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq5_HTML.gif: If their determinants are equal, det P 1 = det P 2 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq6_HTML.gif, then

        tr P 1 tr P 2 and tr Cof P 1 tr Cof P 2 log P 1 F 2 log P 2 F 2 , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equc_HTML.gif

        where log is the principal matrix logarithm and P F 2 = i , j = 1 3 P i j 2 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq7_HTML.gif denotes the Frobenius matrix norm. Applications in matrix analysis and nonlinear elasticity are indicated.

        MSC:26D05, 26D07.

        Keywords

        matrix logarithm elementary symmetric polynomials inequality characteristic polynomial positive definite matrices means

        1 Introduction

        Convexity is a powerful source for obtaining new inequalities; see, e.g., [1, 2]. In applications coming from nonlinear elasticity, we are faced, however, with variants of the squared logarithm function; see the last section. The function ( log ( x ) ) 2 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq8_HTML.gif is neither convex nor concave. Nevertheless, the sum of squared logarithms inequality holds. We will proceed as follows: In the first section, we will give several equivalent formulations of the inequality, for example, in terms of the coefficients of the characteristic polynomial (Theorem 1), in terms of elementary symmetric polynomials (Theorem 3), in terms of means (Theorem 5) or in terms of the Frobenius matrix norm (Theorem 7). A proof of the inequality will be given in Section 2, and some counterexamples for slightly changed variants of the inequality are discussed in Section 3. In the last section, an application of the sum of squared logarithms inequality in matrix analysis and in the mathematical theory of nonlinear elasticity is indicated.

        2 Formulations of the problem

        All theorems in this section are equivalent.

        Theorem 1 For n = 2 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq9_HTML.gif or n = 3 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq10_HTML.gif let P 1 , P 2 R n × n http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq11_HTML.gif be positive definite real matrices. Let the coefficients of the characteristic polynomials of P 1 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq4_HTML.gif and P 2 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq5_HTML.gif satisfy
        tr P 1 tr P 2 and tr Cof P 1 tr Cof P 2 and det P 1 = det P 2 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equd_HTML.gif
        Then
        log P 1 F 2 log P 2 F 2 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Eque_HTML.gif

        For n = 3 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq10_HTML.gif, we will now give equivalent formulations of this statement. The case n = 2 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq9_HTML.gif can be treated analogously. For its proof, see Remark 15. By orthogonal diagonalization of P 1 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq4_HTML.gif and P 2 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq5_HTML.gif, the inequalities can be rewritten in terms of the eigenvalues y 1 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq12_HTML.gif, y 2 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq13_HTML.gif, y 3 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq14_HTML.gif and a 1 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq15_HTML.gif, a 2 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq16_HTML.gif, a 3 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq17_HTML.gif, respectively.

        Theorem 2 Let the real numbers a 1 , a 2 , a 3 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq18_HTML.gif and y 1 , y 2 , y 3 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq19_HTML.gif be such that
        y 1 + y 2 + y 3 a 1 + a 2 + a 3 , y 1 y 2 + y 2 y 3 + y 1 y 3 a 1 a 2 + a 2 a 3 + a 1 a 3 , y 1 y 2 y 3 = a 1 a 2 a 3 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ1_HTML.gif
        (1)
        Then
        ( log y 1 ) 2 + ( log y 2 ) 2 + ( log y 3 ) 2 ( log a 1 ) 2 + ( log a 2 ) 2 + ( log a 3 ) 2 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ2_HTML.gif
        (2)
        The elementary symmetric polynomials, see, e.g., [[3], p.178]
        e 0 ( y 1 , y 2 , y 3 ) = 1 , e 1 ( y 1 , y 2 , y 3 ) = y 1 + y 2 + y 3 , e 2 ( y 1 , y 2 , y 3 ) = y 1 y 2 + y 1 y 3 + y 2 y 3 , e 3 ( y 1 , y 2 , y 3 ) = y 1 y 2 y 3 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equf_HTML.gif

        are known to have the Schur-concavity property (i.e., e k http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq20_HTML.gif is Schur-convex) [1, 4]; see (16). It is possible to express the problem in terms of these elementary symmetric polynomials as follows.

        Theorem 3 Let a 1 , a 2 , a 3 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq18_HTML.gif and y 1 , y 2 , y 3 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq19_HTML.gif satisfy
        e 1 ( y 1 , y 2 , y 3 ) e 1 ( a 1 , a 2 , a 3 ) , e 2 ( y 1 , y 2 , y 3 ) e 2 ( a 1 , a 2 , a 3 ) , e 3 ( y 1 , y 2 , y 3 ) = e 3 ( a 1 , a 2 , a 3 ) . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equg_HTML.gif
        Then
        e 1 ( ( log y 1 ) 2 , ( log y 2 ) 2 , ( log y 3 ) 2 ) e 1 ( ( log a 1 ) 2 , ( log a 2 ) 2 , ( log a 3 ) 2 ) . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equh_HTML.gif
        Because y 1 y 2 y 3 = a 1 a 2 a 3 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq21_HTML.gif, we have
        y 1 y 2 + y 2 y 3 + y 1 y 3 a 1 a 2 + a 2 a 3 + a 1 a 3 1 y 1 + 1 y 2 + 1 y 3 1 a 1 + 1 a 2 + 1 a 3 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equi_HTML.gif

        Thus, we obtain the following theorem.

        Theorem 4 Let the real numbers a 1 , a 2 , a 3 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq18_HTML.gif and y 1 , y 2 , y 3 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq19_HTML.gif be such that
        y 1 + y 2 + y 3 a 1 + a 2 + a 3 , 1 y 1 + 1 y 2 + 1 y 3 1 a 1 + 1 a 2 + 1 a 3 , y 1 y 2 y 3 = a 1 a 2 a 3 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ3_HTML.gif
        (3)
        Then
        ( log y 1 ) 2 + ( log y 2 ) 2 + ( log y 3 ) 2 ( log a 1 ) 2 + ( log a 2 ) 2 + ( log a 3 ) 2 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ4_HTML.gif
        (4)
        The conditions (3) are also simple expressions in terms of arithmetic, harmonic and geometric and quadratic mean
        A ( y 1 , y 2 , y 3 ) = y 1 + y 2 + y 3 3 , H ( y 1 , y 2 , y 3 ) = 3 1 y 1 + 1 y 2 + 1 y 3 , G ( y 1 , y 2 , y 3 ) = y 1 y 2 y 3 3 , Q ( y 1 , y 2 , y 3 ) = 1 3 ( y 1 2 + y 2 2 + y 3 2 ) . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equj_HTML.gif
        Theorem 5 Let a 1 , a 2 , a 3 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq18_HTML.gif and y 1 , y 2 , y 3 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq19_HTML.gif. Then A ( y 1 , y 2 , y 3 ) A ( a 1 , a 2 , a 3 ) http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq22_HTML.gif, H ( a 1 , a 2 , a 3 ) H ( y 1 , y 2 , y 3 ) http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq23_HTML.gif (‘reverse!) and G ( y 1 , y 2 , y 3 ) = G ( a 1 , a 2 , a 3 ) http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq24_HTML.gif imply
        Q ( log y 1 , log y 2 , log y 3 ) Q ( log a 1 , log a 2 , log a 3 ) . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equk_HTML.gif
        We denote by
        a i = : d i 2 , y i = : x i 2 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equl_HTML.gif

        and arrive at

        Theorem 6 Let the real numbers d i http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq25_HTML.gif and x i http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq26_HTML.gif be such that d 1 , d 2 , d 3 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq27_HTML.gif, x 1 , x 2 , x 3 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq28_HTML.gif and
        x 1 2 + x 2 2 + x 3 2 d 1 2 + d 2 2 + d 3 2 , x 1 2 x 2 2 + x 2 2 x 3 2 + x 1 2 x 3 2 d 1 2 d 2 2 + d 2 2 d 3 2 + d 1 2 d 3 2 , x 1 x 2 x 3 = d 1 d 2 d 3 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ5_HTML.gif
        (5)
        Then
        ( log x 1 ) 2 + ( log x 2 ) 2 + ( log x 3 ) 2 ( log d 1 ) 2 + ( log d 2 ) 2 + ( log d 3 ) 2 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ6_HTML.gif
        (6)

        If we again view x i http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq26_HTML.gif and d i http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq25_HTML.gif as eigenvalues of positive definite matrices, an equivalent formulation of the problem can be given in terms of their Frobenius matrix norms:

        Theorem 7 For n { 2 , 3 } http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq29_HTML.gif, let P 1 , P 2 R n × n http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq11_HTML.gif be positive definite real matrices. Let
        P 1 F 2 P 2 F 2 and P 1 1 F 2 P 2 1 F 2 and det P 1 = det P 2 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equm_HTML.gif
        Then
        log P 1 F 2 log P 2 F 2 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equn_HTML.gif
        Let us reconsider the formulation from Theorem 5. If we denote
        c i : = log a i , z i : = log y i , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equo_HTML.gif
        from H ( a 1 , a 2 , a 3 ) H ( y 1 , y 2 , y 3 ) http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq30_HTML.gif, we obtain
        e z 1 + e z 2 + e z 3 e c 1 + e c 2 + e c 3 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equp_HTML.gif
        Theorem 8 Let the real numbers c 1 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq31_HTML.gif, c 2 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq32_HTML.gif, c 3 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq33_HTML.gif and z 1 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq34_HTML.gif, z 2 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq35_HTML.gif, z 3 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq36_HTML.gif be such that
        e z 1 + e z 2 + e z 3 e c 1 + e c 2 + e c 3 , e z 1 + e z 2 + e z 3 e c 1 + e c 2 + e c 3 , z 1 + z 2 + z 3 = c 1 + c 2 + c 3 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ7_HTML.gif
        (7)
        Then
        z 1 2 + z 2 2 + z 3 2 c 1 2 + c 2 2 + c 3 2 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ8_HTML.gif
        (8)
        In order to prove Theorem 8, one can assume without loss of generality that
        z 1 + z 2 + z 3 = c 1 + c 2 + c 3 = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ9_HTML.gif
        (9)

        Thus, we have the equivalent formulation

        Theorem 9 Let the real numbers c ¯ 1 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq37_HTML.gif, c ¯ 2 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq38_HTML.gif, c ¯ 3 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq39_HTML.gif and z ¯ 1 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq40_HTML.gif, z ¯ 2 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq41_HTML.gif, z ¯ 3 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq42_HTML.gif be such that
        e z ¯ 1 + e z ¯ 2 + e z ¯ 3 e c ¯ 1 + e c ¯ 2 + e c ¯ 3 , e z ¯ 1 + e z ¯ 2 + e z ¯ 3 e c ¯ 1 + e c ¯ 2 + e c ¯ 3 , z ¯ 1 + z ¯ 2 + z ¯ 3 = c ¯ 1 + c ¯ 2 + c ¯ 3 = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ10_HTML.gif
        (10)
        Then
        z ¯ 1 2 + z ¯ 2 2 + z ¯ 3 2 c ¯ 1 2 + c ¯ 2 2 + c ¯ 3 2 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ11_HTML.gif
        (11)
        Let us prove that Theorem 8 can be reformulated as Theorem 9. Indeed, let us assume that Theorem 9 is valid and show that the statement of Theorem 8 also holds true. We denote by s the sum s = z 1 + z 2 + z 3 = c 1 + c 2 + c 3 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq43_HTML.gif and we designate
        z ¯ i = z i s 3 , c ¯ i = c i s 3 ( i = 1 , 2 , 3 ) . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equq_HTML.gif
        Then the real numbers z ¯ i http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq44_HTML.gif and c ¯ i http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq45_HTML.gif satisfy the hypotheses of Theorem 9 and we obtain z ¯ 1 2 + z ¯ 2 2 + z ¯ 3 2 c ¯ 1 2 + c ¯ 2 2 + c ¯ 3 2 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq46_HTML.gif. This inequality is equivalent to
        i = 1 3 ( z i s 3 ) 2 i = 1 3 ( c i s 3 ) 2 , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equr_HTML.gif
        which, by virtue of the condition (7)3, reduces to
        z 1 2 + z 2 2 + z 3 2 c 1 2 + c 2 2 + c 3 2 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equs_HTML.gif

        Thus, Theorem 8 is also valid.

        By virtue of the logical equivalence
        ( A B C ) ( ¬ C ¬ A ¬ B ) http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equt_HTML.gif

        for any statements A, B, C, we can formulate the inequality (11) (i.e., Theorem 9) in the following equivalent manner.

        Theorem 10 Let the real numbers c 1 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq31_HTML.gif, c 2 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq32_HTML.gif, c 3 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq33_HTML.gif and z 1 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq34_HTML.gif, z 2 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq35_HTML.gif, z 3 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq36_HTML.gif be such that
        z 1 + z 2 + z 3 = c 1 + c 2 + c 3 = 0 and z 1 2 + z 2 2 + z 3 2 < c 1 2 + c 2 2 + c 3 2 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ12_HTML.gif
        (12)
        Then one of the following inequalities holds:
        e z 1 + e z 2 + e z 3 < e c 1 + e c 2 + e c 3 or e z 1 + e z 2 + e z 3 < e c 1 + e c 2 + e c 3 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ13_HTML.gif
        (13)

        We use the statement of Theorem 10 for the proof.

        Before continuing, let us show that our new inequality is not a consequence of majorization and Karamata’s inequality [5]. Consider z = ( z 1 , , z n ) R + n http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq47_HTML.gif and c = ( c 1 , , c n ) R + n http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq48_HTML.gif arranged already in decreasing order z 1 z 2 z n http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq49_HTML.gif and c 1 c 2 c n http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq50_HTML.gif. If
        i = 1 k z i i = 1 k c i ( 1 k n 1 ) , i = 1 n z i = i = 1 n c i , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ14_HTML.gif
        (14)
        we say that z majorizes c, denoted by z c http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq51_HTML.gif. The following result is well known [[6], p.89], [4, 5]. If f : R R http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq52_HTML.gif is convex, then
        z c i = 1 n f ( z i ) i = 1 n f ( c i ) . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ15_HTML.gif
        (15)
        A function g : R n R http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq53_HTML.gif which satisfies
        z c g ( z 1 , , z n ) g ( c 1 , , c n ) http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ16_HTML.gif
        (16)
        is called Schur-convex. In Theorem 8, the convex function to be considered would be f ( t ) = t 2 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq54_HTML.gif. Do conditions (7) (upon rearrangement of z , c R + 3 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq55_HTML.gif if necessary) yield already majorization z c http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq51_HTML.gif? This is not the case, as we explain now. Let the real numbers z 1 z 2 z 3 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq56_HTML.gif and c 1 c 2 c 3 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq57_HTML.gif be such that
        e z 1 + e z 2 + e z 3 e c 1 + e c 2 + e c 3 , e z 1 + e z 2 + e z 3 e c 1 + e c 2 + e c 3 , z 1 + z 2 + z 3 = c 1 + c 2 + c 3 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ17_HTML.gif
        (17)
        These conditions do not imply the majorization z c http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq51_HTML.gif,
        z 1 c 1 , z 1 + z 2 c 1 + c 2 , z 1 + z 2 + z 3 = c 1 + c 2 + c 3 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ18_HTML.gif
        (18)

        Therefore, our inequality (i.e., z 1 2 + z 2 2 + z 3 2 c 1 2 + c 2 2 + c 3 2 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq58_HTML.gif) does not follow from majorization in disguise.

        Indeed, let
        z 1 = 1 2 + 0.95 2 3 , z 2 = 1 2 + 0.85 2 3 , z 3 = 1 0.9 3 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equu_HTML.gif
        and
        c 1 = 1 2 + 1 2 3 , c 2 = 1 2 + 1 2 3 , c 3 = 1 3 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equv_HTML.gif
        Then we have z 1 > z 2 > z 3 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq59_HTML.gif and c 1 > c 2 > c 3 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq60_HTML.gif, together with
        e z 1 + e z 2 + e z 3 = 4.49497 > 3.57137 = e c 1 + e c 2 + e c 3 , e z 1 + e z 2 + e z 3 = 5.50607 > 3.47107 = e c 1 + e c 2 + e c 3 , z 1 + z 2 + z 3 = c 1 + c 2 + c 3 = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equw_HTML.gif

        but the majorization inequalities (18) are not satisfied, since z 1 < c 1 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq61_HTML.gif.

        3 Proof of the inequality

        Of course, we may assume without loss of generality that c 1 c 2 c 3 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq57_HTML.gif and z 1 z 2 z 3 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq56_HTML.gif (and the same for a i http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq62_HTML.gif, d i http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq25_HTML.gif, x i http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq26_HTML.gif, y i http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq63_HTML.gif).

        The proof begins with the crucial lemma.

        Lemma 11 Let the real numbers a b c http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq64_HTML.gif and x y z http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq65_HTML.gif be such that
        a + b + c = x + y + z = 0 , a 2 + b 2 + c 2 = x 2 + y 2 + z 2 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ19_HTML.gif
        (19)
        Then the inequality
        e a + e b + e c e x + e y + e z http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ20_HTML.gif
        (20)
        is satisfied if and only if the relation
        a x http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ21_HTML.gif
        (21)
        holds, or equivalently, if and only if
        c z http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ22_HTML.gif
        (22)

        holds.

        Proof Let us denote by r : = 2 3 ( a 2 + b 2 + c 2 ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq66_HTML.gif. Then, from (19), it follows
        b + c = a , b 2 + c 2 = 3 2 r 2 a 2 , y + z = x , y 2 + z 2 = 3 2 r 2 x 2 , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equx_HTML.gif
        and we find
        b = 1 2 ( a + 3 ( r 2 a 2 ) ) , c = 1 2 ( a 3 ( r 2 a 2 ) ) , y = 1 2 ( x + 3 ( r 2 x 2 ) ) , z = 1 2 ( x 3 ( r 2 x 2 ) ) . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ23_HTML.gif
        (23)
        In view of (19) and a b c http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq67_HTML.gif, x y z http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq68_HTML.gif, one can show that
        a , x [ r 2 , r ] , b , y [ r 2 , r 2 ] , c , z [ r , r 2 ] . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ24_HTML.gif
        (24)
        Indeed, let us verify the relations (24). We have
        r 2 a r 1 6 ( a 2 + b 2 + c 2 ) a 2 2 3 ( a 2 + b 2 + c 2 ) b 2 + c 2 5 a 2 and a 2 2 ( b 2 + c 2 ) b 2 + ( a + b ) 2 5 a 2 and ( b + c ) 2 2 ( b 2 + c 2 ) 4 a 2 2 a b 2 b 2 0 and b 2 + c 2 2 b c 2 ( a b ) ( 2 a + b ) 0 and ( b c ) 2 0 , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equy_HTML.gif
        which hold true since a b http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq69_HTML.gif and 2 a + b a + b + c = 0 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq70_HTML.gif. Similarly, we have
        r 2 b r 2 b 2 r 2 4 4 b 2 2 3 ( a 2 + b 2 + c 2 ) 5 b 2 a 2 + c 2 5 b 2 a 2 + ( a + b ) 2 2 a 2 + 2 a b 4 b 2 0 2 ( a b ) ( a + 2 b ) 0 , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equz_HTML.gif
        which holds true since a b http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq69_HTML.gif and a + 2 b a + b + c = 0 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq71_HTML.gif. Also, we have
        r c r 2 r 2 c 2 r 2 4 2 3 ( a 2 + b 2 + c 2 ) c 2 1 6 ( a 2 + b 2 + c 2 ) 2 ( a 2 + b 2 ) c 2 and 5 c 2 a 2 + b 2 2 ( a 2 + b 2 ) ( a + b ) 2 and 5 ( a + b ) 2 a 2 + b 2 ( a b ) 2 0 and 4 a 2 + 10 a b + 4 b 2 0 ( a b ) 2 0 and 2 ( a + 2 b ) ( 2 a + b ) 0 , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equaa_HTML.gif

        which hold true since a + 2 b a + b + c = 0 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq71_HTML.gif and 2 a + b a + b + c = 0 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq70_HTML.gif. One can show in the same way that x [ r 2 , r ] http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq72_HTML.gif, y [ r 2 , r 2 ] http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq73_HTML.gif, z [ r , r 2 ] http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq74_HTML.gif, so that (24) has been verified.

        We prove now that the inequality (21) holds if and only if (22) holds. Indeed, using (23)2,4 and (24) we get
        c z a 3 ( r 2 a 2 ) x 3 ( r 2 x 2 ) a r + 3 ( 1 ( a r ) 2 ) x r + 3 ( 1 ( x r ) 2 ) a x , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equab_HTML.gif

        since the function t t + 3 ( 1 t 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq75_HTML.gif is decreasing for t [ 1 2 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq76_HTML.gif.

        Let us prove next that the inequalities (20) and (21) are equivalent. To accomplish this, we introduce the function f : [ r 2 , r ] R http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq77_HTML.gif by
        f ( x ) = e x + e ( x + 3 ( r 2 x 2 ) ) / 2 + e ( x 3 ( r 2 x 2 ) ) / 2 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ25_HTML.gif
        (25)
        Taking into account (23) and (24)1, the inequality (20) can be written equivalently as
        f ( a ) f ( x ) , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ26_HTML.gif
        (26)
        which is equivalent to
        a x , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equac_HTML.gif
        since the function f defined by (25) is monotone increasing on [ r 2 , r ] http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq78_HTML.gif, as we show next. To this aim, we denote by
        cos φ : = x r [ 1 2 , 1 ] , i.e. φ : = arccos ( x r ) [ 0 , π 3 ] . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equad_HTML.gif
        Then the function (25) can be written as
        f ( x ) = h ( r , φ ) , where  h : ( 0 , ) × [ 0 , π 3 ] R , h ( r , φ ) = e r cos φ + e r cos ( φ + 2 π / 3 ) + e r cos ( φ 2 π / 3 ) . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ27_HTML.gif
        (27)
        We have to show that h ( r , φ ) http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq79_HTML.gif is decreasing with respect to φ [ 0 , π 3 ] http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq80_HTML.gif. We compute the first derivative
        φ h ( r , φ ) = r [ e r cos φ sin φ + e r cos ( φ + 2 π / 3 ) sin ( φ + 2 π 3 ) + e r cos ( φ 2 π / 3 ) sin ( φ 2 π 3 ) ] . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ28_HTML.gif
        (28)
        The function (28) has the same sign as the function
        F ( r , φ ) : = 1 r e r cos φ φ h ( r , φ ) , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ29_HTML.gif
        (29)
        i.e., the function F : ( 0 , ) × [ 0 , π 3 ] R http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq81_HTML.gif given by
        F ( r , φ ) = sin φ e r 3 sin ( φ + π / 3 ) sin ( φ + 2 π 3 ) e r 3 sin ( φ π / 3 ) sin ( φ 2 π 3 ) . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ30_HTML.gif
        (30)
        In order to show that F ( r , φ ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq82_HTML.gif for all ( r , φ ) ( 0 , ) × [ 0 , π 3 ] http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq83_HTML.gif, we remark that lim r 0 F ( r , φ ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq84_HTML.gif for fixed φ [ 0 , π 3 ] http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq85_HTML.gif and we compute
        r F ( r , φ ) = 3 [ e r 3 sin ( φ + π / 3 ) sin ( φ + π 3 ) sin ( φ + 2 π 3 ) e r 3 sin ( φ π / 3 ) sin ( φ π 3 ) sin ( φ 2 π 3 ) ] = 3 [ e r 3 sin ( φ + π / 3 ) 1 2 ( cos ( 2 φ + π ) + cos π 3 ) e r 3 sin ( φ π / 3 ) 1 2 ( cos ( 2 φ π ) + cos π 3 ) ] = 3 2 ( cos 2 φ + 1 2 ) [ e r 3 sin ( φ + π / 3 ) e r 3 sin ( φ π / 3 ) ] 0 , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equae_HTML.gif

        since φ [ 0 , π 3 ] http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq86_HTML.gif implies cos 2 φ 1 2 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq87_HTML.gif and sin ( φ + π 3 ) sin ( φ π 3 ) http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq88_HTML.gif.

        Consequently, the function F ( r , φ ) http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq89_HTML.gif is decreasing with respect to r and for any ( r , φ ) ( 0 , ) × [ 0 , π 3 ] http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq90_HTML.gif we have that
        F ( r , φ ) lim r 0 F ( r , φ ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ31_HTML.gif
        (31)

        From (29) and (31), it follows that h ( r , φ ) http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq91_HTML.gif is decreasing with respect to φ [ 0 , π 3 ] http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq92_HTML.gif. This means that f ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq93_HTML.gif is increasing as a function of x [ r 2 , r ] http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq72_HTML.gif, i.e., the relation (26) is indeed equivalent to a x http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq94_HTML.gif and the proof is complete. □

        Consequence 12 Let the real numbers a b c http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq67_HTML.gif and x y z http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq65_HTML.gif be such that
        a + b + c = x + y + z = 0 , a 2 + b 2 + c 2 = x 2 + y 2 + z 2 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equaf_HTML.gif
        Then one of the following inequalities holds:
        e a + e b + e c e x + e y + e z , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ32_HTML.gif
        (32)
        or
        e a + e b + e c e x + e y + e z . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ33_HTML.gif
        (33)

        The inequalities (32) and (33) are satisfied simultaneously if and only if a = x http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq95_HTML.gif, b = y http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq96_HTML.gif and c = z http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq97_HTML.gif.

        Proof According to Lemma 11, the inequality (32) is equivalent to
        a x , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ34_HTML.gif
        (34)
        while the inequality (33) is equivalent to
        a x . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ35_HTML.gif
        (35)

        Since one of the relations (34) and (35) must hold, we have proved that one of the inequalities (32) and (33) is satisfied. They are simultaneously satisfied if and only if both (34) and (35) hold true, i.e., a = x http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq95_HTML.gif (and consequently b = y http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq96_HTML.gif, c = z http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq97_HTML.gif). □

        Consequence 13 Let the real numbers a b c http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq67_HTML.gif and x y z http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq65_HTML.gif be such that
        a + b + c = x + y + z = 0 , a 2 + b 2 + c 2 = x 2 + y 2 + z 2 and e a + e b + e c = e x + e y + e z . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equag_HTML.gif

        Then we have a = x http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq95_HTML.gif, b = y http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq96_HTML.gif and c = z http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq97_HTML.gif.

        Proof Since by hypothesis e a + e b + e c e x + e y + e z http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq98_HTML.gif holds, we can apply Lemma 11 to deduce a x http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq99_HTML.gif and c z http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq100_HTML.gif.

        On the other hand, by virtue of the inverse inequality e x + e y + e z e a + e b + e c http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq101_HTML.gif and Lemma 11, we obtain x a http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq102_HTML.gif and z c http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq103_HTML.gif. In conclusion, we get a = x http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq95_HTML.gif, c = z http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq97_HTML.gif and b = y http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq96_HTML.gif. □

        Proof of Theorem 10 In order to prove (13), we define the real numbers
        t i = k z i ( i = 1 , 2 , 3 ) , where  k = c 1 2 + c 2 2 + c 3 2 z 1 2 + z 2 2 + z 3 2 > 1 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ36_HTML.gif
        (36)
        Then we have
        t 1 + t 2 + t 3 = c 1 + c 2 + c 3 = 0 and t 1 2 + t 2 2 + t 3 2 = c 1 2 + c 2 2 + c 3 2 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ37_HTML.gif
        (37)
        If we apply the Consequence 12 for the numbers c 1 c 2 c 3 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq57_HTML.gif and t 1 t 2 t 3 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq104_HTML.gif, then we obtain that
        e t 1 + e t 2 + e t 3 e c 1 + e c 2 + e c 3 or e t 1 + e t 2 + e t 3 e c 1 + e c 2 + e c 3 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ38_HTML.gif
        (38)
        In what follows, let us show that
        e z 1 + e z 2 + e z 3 < e t 1 + e t 2 + e t 3 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ39_HTML.gif
        (39)
        Using the notations ρ : = 2 3 ( z 1 2 + z 2 2 + z 3 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq105_HTML.gif and
        cos ζ : = z 1 ρ [ 1 2 , 1 ] , i.e. ζ : = arccos ( z 1 ρ ) [ 0 , π 3 ] , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equah_HTML.gif
        we have k ρ : = 2 3 ( t 1 2 + t 2 2 + t 3 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq106_HTML.gif and cos ζ = t 1 k ρ http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq107_HTML.gif. With the help of the function h defined in (27), we can write the inequality (39) in the form
        e ρ cos ζ + e ρ cos ( ζ + 2 π / 3 ) + e ρ cos ( ζ 2 π / 3 ) < e k ρ cos ζ + e k ρ cos ( ζ + 2 π / 3 ) + e k ρ cos ( ζ 2 π / 3 ) , or h ( ρ , ζ ) < h ( k ρ , ζ ) , ( ρ , ζ ) ( 0 , ) × [ 0 , π 3 ] , k > 1 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ40_HTML.gif
        (40)
        The relation (40) asserts that the function h defined in (27) is increasing with respect to the first variable r ( 0 , ) http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq108_HTML.gif. To show this, we compute the derivative
        r h ( r , φ ) = e r cos φ cos φ + e r cos ( φ + 2 π / 3 ) cos ( φ + 2 π 3 ) + e r cos ( φ 2 π / 3 ) cos ( φ 2 π 3 ) . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ41_HTML.gif
        (41)
        By virtue of the Chebyshev’s sum inequality, we deduce from (41) that
        r h ( r , φ ) > 0 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ42_HTML.gif
        (42)
        Indeed, the Chebyshev’s sum inequality [[6], 2.17] asserts that: if a 1 a 2 a n http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq109_HTML.gif and b 1 b 2 b n http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq110_HTML.gif then
        n k = 1 n a k b k ( k = 1 n a k ) ( k = 1 n b k ) . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equai_HTML.gif
        In our case, we derive the following result: for any real numbers x, y, z such that x + y + z = 0 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq111_HTML.gif, the inequality
        x e x + y e y + z e z 1 3 ( x + y + z ) ( e x + e y + e z ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ43_HTML.gif
        (43)

        holds true, with equality if and only if x = y = z = 0 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq112_HTML.gif.

        Applying the result (43) to the function (41), we deduce the relation (42). This means that h ( r , φ ) http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq91_HTML.gif is an increasing function of r, i.e. the inequality (40) holds, and hence, we have proved (39).

        One can show analogously that the inequality
        e z 1 + e z 2 + e z 3 < e t 1 + e t 2 + e t 3 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ44_HTML.gif
        (44)

        is also valid. From (38), (39) and (44), it follows that the assertion (13) holds true. Thus, the proof of Theorem 10 is complete. □

        Since the statements of the Theorems 8 and 10 are equivalent, we have proved also the inequality (8).

        Remark 14 The inequality (8) becomes an equality if and only if z i = c i http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq113_HTML.gif, i = 1 , 2 , 3 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq114_HTML.gif.

        Proof Indeed, assume that z 1 2 + z 2 2 + z 3 2 = c 1 2 + c 2 2 + c 3 2 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq115_HTML.gif. Then we can apply the Consequence 12 and we deduce that
        e z 1 + e z 2 + e z 3 e c 1 + e c 2 + e c 3 or e z 1 + e z 2 + e z 3 e c 1 + e c 2 + e c 3 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ45_HTML.gif
        (45)
        Taking into account (7)1,2 in conjunction with (45), we find
        e z 1 + e z 2 + e z 3 = e c 1 + e c 2 + e c 3 or e z 1 + e z 2 + e z 3 = e c 1 + e c 2 + e c 3 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ46_HTML.gif
        (46)

        By virtue of (46), we can apply the Consequence 13 to derive z 1 = c 1 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq116_HTML.gif, and consequently z 2 = c 2 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq117_HTML.gif, z 3 = c 3 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq118_HTML.gif. □

        Let us prove the following version of the inequality (6) for two pairs of numbers d 1 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq119_HTML.gif, d 2 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq120_HTML.gif and x 1 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq121_HTML.gif, x 2 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq122_HTML.gif:

        Remark 15 If the real numbers d 1 d 2 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq123_HTML.gif and x 1 x 2 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq124_HTML.gif are such that
        x 1 2 + x 2 2 d 1 2 + d 2 2 and x 1 x 2 = d 1 d 2 = 1 , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ47_HTML.gif
        (47)
        then the inequality
        ( log x 1 ) 2 + ( log x 2 ) 2 ( log d 1 ) 2 + ( log d 2 ) 2 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ48_HTML.gif
        (48)
        holds true. Note that the additional condition
        1 x 1 2 + 1 x 2 2 1 d 1 2 + 1 d 2 2 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equaj_HTML.gif

        is automatically fulfilled.

        Proof Since x 1 x 2 = d 1 d 2 = 1 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq125_HTML.gif and d 1 d 2 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq123_HTML.gif, x 1 x 2 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq124_HTML.gif, we have x 1 1 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq126_HTML.gif, d 1 1 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq127_HTML.gif and
        log x 1 = log x 2 0 , log d 1 = log d 2 0 , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equak_HTML.gif

        so that the inequality (48) is equivalent to log x 1 log d 1 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq128_HTML.gif, i.e., we have to show that x 1 d 1 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq129_HTML.gif.

        Indeed, if we insert x 2 = 1 x 1 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq130_HTML.gif and d 2 = 1 d 1 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq131_HTML.gif into the inequality (47)1 then we find
        x 1 2 + 1 x 1 2 d 1 2 + 1 d 1 2 , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equal_HTML.gif

        which means that x 1 d 1 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq132_HTML.gif since the function t t 2 + 1 t 2 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq133_HTML.gif is increasing for t [ 1 , ) http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq134_HTML.gif. This completes the proof. □

        Alternative proof of Remark 15 Let x 3 = d 3 = 1 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq135_HTML.gif. Then (47) implies x 1 2 + x 2 2 + x 3 2 d 1 2 + d 2 2 + d 3 2 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq136_HTML.gif and x 1 x 2 x 3 = d 1 d 2 d 3 = 1 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq137_HTML.gif as well as
        x 1 2 x 2 2 + x 2 2 x 3 2 + x 1 2 x 3 2 = 1 + x 2 2 + x 1 2 1 + d 2 2 + d 1 2 = d 1 2 d 2 2 + d 2 2 d 3 2 + d 1 2 d 3 2 , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ49_HTML.gif
        (49)

        because x 1 2 x 2 2 = 1 = d 1 2 d 2 2 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq138_HTML.gif, and Theorem 6 provides the assertion. □

        4 Some counterexamples for weakened assumptions

        Example 16 Unlike in the 2D case in Remark 15, for two triples of numbers the second condition (18)2 of Theorem 2, namely y 1 y 2 + y 2 y 3 + y 1 y 3 a 1 a 2 + a 2 a 3 + a 1 a 3 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq139_HTML.gif, cannot be removed. Let
        y 1 = e 6 , y 2 = 1 , y 3 = e 6 , a 1 = e 4 , a 2 = e 4 , a 3 = e 8 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equam_HTML.gif
        Then y 1 y 2 y 3 = a 1 a 2 a 3 = 1 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq140_HTML.gif and
        y 1 + y 2 + y 3 > e 6 e 2 e 4 > 3 e 4 > a 1 + a 2 + a 3 , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equan_HTML.gif
        but
        ( log y 1 ) 2 + ( log y 2 ) 2 + ( log y 3 ) 2 = 36 + 0 + 36 < 16 + 16 + 64 = ( log a 1 ) 2 + ( log a 2 ) 2 + ( log a 3 ) 2 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equao_HTML.gif
        Example 17 The condition y 1 y 2 y 3 = a 1 a 2 a 3 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq141_HTML.gif cannot be weakened to y 1 y 2 y 3 a 1 a 2 a 3 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq142_HTML.gif. Indeed, let y 2 = y 3 = a 1 = a 2 = 1 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq143_HTML.gif, y 1 = e http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq144_HTML.gif, a 3 = e 2 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq145_HTML.gif. Then
        y 1 + y 2 + y 3 = e + 1 + 1 1 + 1 + e 2 = a 1 + a 2 + a 3 , y 1 y 2 + y 1 y 3 + y 2 y 3 = e + e + 1 1 + e 2 + e 2 = a 1 a 2 + a 1 a 3 + a 2 a 3 , y 1 y 2 y 3 = e e 2 = a 1 a 2 a 3 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equap_HTML.gif
        But nevertheless
        ( log y 1 ) 2 + ( log y 2 ) 2 + ( log y 3 ) 2 = 1 + 0 + 0 < 0 + 0 + 4 = ( log a 1 ) 2 + ( log a 2 ) 2 + ( log a 3 ) 2 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equaq_HTML.gif

        A counterexample for the two variable case can be constructed analogously.

        Example 18 Even with an analogous condition, the inequality (4) does not hold for n = 4 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq146_HTML.gif numbers (without further assumptions). Indeed, let
        y 1 = e , y 2 = y 3 = e 7 , y 4 = e 15 , a 1 = a 2 = e 6 , a 3 = e 7 , a 4 = e 19 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equar_HTML.gif
        Then y 1 y 2 y 3 y 4 = a 1 a 2 a 3 a 4 = 1 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq147_HTML.gif. Also,
        y 1 + y 2 + y 3 + y 4 = e + e 7 + e 7 + e 15 > 0 + e 7 + 2 e 6 + e 19 = a 1 + a 2 + a 3 + a 4 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equas_HTML.gif
        Furthermore,
        y 1 y 2 + y 1 y 3 + y 1 y 4 + y 2 y 3 + y 2 y 4 + y 3 y 4 = e 8 + e 8 + e 14 + e 14 + e 8 + e 8 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equat_HTML.gif
        and
        a 1 a 2 + a 1 a 3 + a 1 a 4 + a 2 a 3 + a 2 a 4 + a 3 a 4 = e 12 + e 13 + e 13 + e 13 + e 13 + e 12 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equau_HTML.gif
        Since e 2 > 2 e + 1 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq148_HTML.gif, we have e 14 > e 13 + e 13 + e 12 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq149_HTML.gif and, therefore,
        y 1 y 2 + y 1 y 3 + y 1 y 4 + y 2 y 3 + y 2 y 4 + y 3 y 4 a 1 a 2 + a 1 a 3 + a 1 a 4 + a 2 a 3 + a 2 a 4 + a 3 a 4 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equav_HTML.gif
        Nevertheless, for the sum of squared logarithms, the ‘reverse’ inequality
        ( log y 1 ) 2 + ( log y 2 ) 2 + ( log y 3 ) 2 + ( log y 4 ) 2 = 1 + 49 + 49 + 225 = 324 < 482 = 36 + 36 + 49 + 361 = ( log a 1 ) 2 + ( log a 2 ) 2 + ( log a 3 ) 2 + ( log a 4 ) 2 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equaw_HTML.gif

        holds true.

        Example 19 The inequality (4) does not remain true either, if the function log ( y ) http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq150_HTML.gif is replaced by its linearization ( y 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq151_HTML.gif. Indeed, let y 1 = 9 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq152_HTML.gif, y 2 = 5 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq153_HTML.gif, y 3 = 1 45 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq154_HTML.gif, a 1 = 10 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq155_HTML.gif, a 2 = 1 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq156_HTML.gif, a 3 = 1 10 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq157_HTML.gif. Then
        y 1 + y 2 + y 3 > 14 > 11.1 = a 1 + a 2 + a 3 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equax_HTML.gif
        and
        y 1 y 2 + y 1 y 3 + y 2 y 3 > 45 11.1 = a 1 a 2 + a 1 a 3 + a 2 a 3 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equay_HTML.gif
        But
        ( y 1 1 ) 2 + ( y 2 1 ) 2 + ( y 3 1 ) 2 = 64 + 16 + ( 44 45 ) 2 < 81 < 9 2 + 0 + ( 9 10 ) 2 = ( a 1 1 ) 2 + ( a 2 1 ) 2 + ( a 3 1 ) 2 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equaz_HTML.gif

        5 Conjecture for arbitrary n

        The structure of the inequality in dimensions n = 2 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq9_HTML.gif and n = 3 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq10_HTML.gif and extensive numerical sampling strongly suggest that the inequality holds for all n N http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq158_HTML.gif if the n corresponding conditions are satisfied. More precisely, in terms of the elementary symmetric polynomials, we expect the following:

        Conjecture 20 Let n N http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq158_HTML.gif and y i , a i > 0 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq159_HTML.gif for i = 1 , , n http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq160_HTML.gif. If for all i = 1 , , n 1 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq161_HTML.gif we have
        e i ( y 1 , , y n ) e i ( a 1 , , a n ) and e n ( y 1 , , y n ) = e n ( a 1 , , a n ) , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equba_HTML.gif
        then
        i = 1 n ( log y i ) 2 i = 1 n ( log a i ) 2 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equbb_HTML.gif

        6 Applications

        The investigation in this paper has been motivated by some recent applications. The new sum of squared logarithms inequality is one of the fundamental tools in deducing a novel optimality result in matrix analysis and the conditions in the form (3) had been deduced in the course of that work. Optimality in the matrix problem suggested the sum of squared logarithms inequality. Indeed, based on the present result in [7], it has been shown that for all invertible Z C 3 × 3 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq162_HTML.gif and for any definition of the matrix logarithm as possibly multivalued solution X C 3 × 3 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq163_HTML.gif of exp X = Z http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq164_HTML.gif it holds
        min Q Q = I log Q Z F 2 = log U p Z F 2 = log H F 2 , min Q Q = I sym log Q Z F 2 = sym log U p Z F 2 = log H F 2 , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ50_HTML.gif
        (50)
        where sym X = 1 2 ( X + X ) http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq165_HTML.gif is the Hermitian part of X C 3 × 3 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq166_HTML.gif and U p http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq167_HTML.gif is the unitary factor in the polar decomposition of Z into unitary and Hermitian positive definite matrix H
        Z = U p H . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ51_HTML.gif
        (51)
        This result (50) generalizes the fact that for any complex logarithm and for all z C { 0 } http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq168_HTML.gif
        min ϑ ( π , π ] | log C [ e i ϑ z ] | 2 = | log R | z | | 2 , min ϑ ( π , π ] | Re log C [ e i ϑ z ] | 2 = | log R | z | | 2 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ52_HTML.gif
        (52)
        The optimality result (50) can now also be viewed as another characterization of the unitary factor in the polar decomposition. In addition, in a forthcoming contribution [8], we use (50) to calculate the geodesic distance of the isochoric part of the deformation gradient F det F 1 3 SL ( 3 , R ) http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq169_HTML.gif to SO ( 3 , R ) http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq170_HTML.gif in the canonical left-invariant Riemannian metric on SL ( 3 , R ) http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq171_HTML.gif, to the effect that
        dist geod 2 ( F det F 1 3 , SO ( 3 , R ) ) = dev 3 log F T F F 2 , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_Equ53_HTML.gif
        (53)

        where dev 3 X = X 1 3 ( tr X ) I http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq172_HTML.gif is the orthogonal projection of X R 3 × 3 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq173_HTML.gif to trace free matrices. Thereby, we provide a rigorous geometric justification for the preferred use of the Hencky-strain measure log F T F F 2 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-168/MediaObjects/13660_2013_Article_623_IEq174_HTML.gif in nonlinear elasticity and plasticity theory [9].

        Declarations

        Acknowledgements

        The first author (MB) was supported by the German state grant: ‘Programm des Bundes und der Länder für bessere Studienbedingungen und mehr Qualität in der Lehre’.

        Authors’ Affiliations

        (1)
        Lehrstuhl für Nichtlineare Analysis und Modellierung, Fakultät für Mathematik, Universität Duisburg-Essen
        (2)
        Department of Mathematics, University ‘A.I. Cuza’ of Iaşi

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        Copyright

        © Bîrsan et al.; licensee Springer. 2013

        This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://​creativecommons.​org/​licenses/​by/​2.​0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.