Skip to main content

The proof of three power-exponential inequalities

Abstract

In this paper we prove three power-exponential inequalities for positive real numbers. In particular, we conclude that this proofs give affirmatively answers to three, until now, open problems (Conjectures 4.4, 2.1 and 2.2) posed by Cîrtoaje (J. Inequal. Pure Appl. Math. 10:21, 2009; J. Nonlinear Sci. Appl. 4(2):130-137, 2011). Moreover, we present a new proof of the inequality a r a + b r b a r b + b r a for all positive real numbers a and b and r[0,e]. In addition, three new conjectures are presented.

1 Introduction

The power-exponential functions have useful applications in mathematical analysis and in other theories like statistics [1], biology [2, 3], optimization [4], ordinary differential equations [5], and probability [6]. In recent years there has been intensive research in this area; see for instance [717] and the recent overview on general mathematical inequalities done by Cerone and Dragomir [18]. Some problems look like very simple but are difficult. For instance, we have the following two classical problems: find the solution of the equation z e z =a and the basic problem of comparing a b and b a for all positive real numbers a and b. The first problem is perhaps one of the most ancient and useful problems concerning to power-exponential functions; see for instance [1921]. It was introduced by Lambert in [22] and has been studied by recognized mathematicians like Euler, Pólya, Szegö, and Knuth; see [2325]. The solution to the problem has inspired the definition of the well-known W-Lambert function; see [26]. For the solution to the second problem, see the discussion given in [27, 28] and more recently in [16]. Moreover, in spite of its algebraic simplicity, both problems are the central topic of a large number of research papers in the last years (see [7, 11, 13] and references therein). In particular, in this paper, we are interested in some inequalities conjectured by Cîrtoaje in [12, 29], which are very close to the second problem. To be more specific, we start by recalling that in [30] was introduced and probed the following assertion: the inequality a a + b b a b + b a holds for all positive real numbers less than or equal to 1. After that, Cîrtoaje [12] introduced, proved and conjectured several results about inequalities for power-exponential functions. In particular, in [12], it was established that the inequality

a r a + b r b a r b + b r a
(1.1)

holds true for r[0,e] and for either ab1/e or 1/eab>0. However, in [12], Cîrtoaje leaves as an open problem the proof of (1.1) for 1>a>1/e>b>0. Moreover, in [12] the following conjectures were introduced:

Conjecture 4.3. If a, b, c are positive real numbers, then a 2 a + b 2 b + c 2 c a 2 b + b 2 c + c 2 a .

Conjecture 4.4. Let r be a positive real number. The inequality

a r a + b r b + c r c a r b + b r c + c r a
(1.2)

holds true for all positive real numbers a, b, c with abc if and only if re.

Conjecture 4.6. Let r be a positive real number. The inequality a r a + b r a 2 holds for all nonnegative real numbers a and b if and only if r3.

Conjecture 4.7. If a and b are nonnegative real numbers such that a+b=2, then a 3 b + b 3 a + 2 4 ( a b ) 4 2.

Conjecture 4.8. If a and b are nonnegative real numbers such that a+b=1, then a 2 b + b 2 a 1.

Afterwards, the analysis of (1.1) was completed by Manyama in [14]. Thereafter, of the Cîrtoaje conjectures, the milestones of the history are the works of Coronel and Huancas [13], Matejíčka [31], Li [9] and Hisasue [10] (see also the work of Cîrtoaje [29]), where they proved Conjectures 4.3, 4.6, 4.7, and 4.8, respectively. Here, we should be comment that the proof of Conjecture 4.4 is still open. Subsequently, in 2011 Cîrtoaje introduced a new proof of (1.1) and presented the following three new conjectures:

Conjecture 2.1. If a,b]0,1] and r]0,e], then 2 a r a b r b a r b + b r a .

Conjecture 2.2. If a,b,c]0,1], then 3 a a b b c c ( a b c ) a + ( a b c ) b + ( a b c ) c .

Conjecture 5.1. If a, b are nonnegative real numbers satisfying a+b=1, and if k1, then a ( 2 b ) k + b ( 2 a ) k 1.

Recently, Miyagi and Nishizawa [7] have proved Conjecture 5.1. However, Conjectures 2.1 and 2.2 are still open. Thus, the main focus of this paper are the proofs of Conjectures 4.4, 2.1, and 2.2.

The main contribution of the present paper is the development of the proof of the following four theorems:

Theorem 1.1 The inequality (1.1) holds, for all positive real numbers a, b and for all r[0,e].

Theorem 1.2 The inequality (1.2) holds, for all positive real numbers a, b, c and for all r[0,e].

Theorem 1.3 The inequality

2 a r a b r b a r b + b r a
(1.3)

holds, for all positive real numbers a, b and for all r[0,e].

Theorem 1.4 Let nN and x i ] 0 , 1 ] n . Then the inequality

n i = 1 n x i x i i = 1 n ( j = 1 n x j ) x i
(1.4)

holds.

Note that Conjectures 4.4, 2.1, and 2.2 are solved by Theorems 1.2, 1.3, 1.4, respectively. Moreover, we develop a proof of Theorem 1.1 which is an alternative proof of (1.1) for all positive real numbers a, b, and r[0,e], which is distinct from the existing proofs given in [14, 29].

The rest of the paper is organized in two sections: In Section 2 we present the proofs of Theorems 1.1, 1.2, 1.3 and 1.4 and in Section 3 we present some remarks and three new conjectures.

2 Proofs of main results

In this section we present the proofs of Theorems 1.1, 1.2, 1.3, 1.4. Firstly, we recall a result of [13]. Then we present the corresponding proofs.

2.1 A preliminary result

For completeness and self-contained structure of the proofs of Theorems 1.1 and 1.2, we need the following result of [13].

Proposition 2.1 Consider s R + with s1, m R + and f,g: R + R defined as follows:

f(t)= t s t γ s +γandg(t)= { e ln ( t ) / ( t 1 ) , t { 0 , 1 } , e 1 , t = 1 , 0 , t = 0 .

Then the following properties are satisfied:

  1. (i)

    f(γ)=0 and f(0)=f(1)= γ s +γ.

  2. (ii)

    If s>1, f is strictly increasing on ]g(s),[ and strictly decreasing on ]0,g(s)[.

  3. (iii)

    If s]0,1[, f is strictly decreasing on ]g(s),[ and strictly increasing on ]0,g(s)[.

  4. (iv)

    g is continuous on R + {0} and strictly increasing on R + . Furthermore y=1 is a horizontal asymptote of y=g(t).

2.2 Proof of Theorem 1.1

Without loss of generality, we assume that a>b. Indeed, we find the proof (1.1) by application of Proposition 2.1 with t= a r b , γ= b r b , and s=a/b. Indeed, we distinguish three cases

(a2) Case a>b>1 (t>γ>1 and s>1). By Proposition 2.1(iv), we note that g(s)<1. Then, by the strictly increasing behavior of f (Proposition 2.1(ii)) we deduce the inequality since:

t>γ>1>g(s),s>1f(t)= a r a a r b b r a + b r b >f(γ)=0.

(b2) Case a>1b (t>1γ and s>1). For γ[g(s),1], we conclude the inequality by almost identical arguments to that used before in (i), since t>1γg(s) and s>1. Otherwise, if γ[0,g(s)], we deduce that

f(t)= a r a a r b b r a + b r b >f(1)=f(0)>f(γ)=0,

which implies the desired inequality.

(c2) Case 1>a>b>0 (1>t>γ>0 and s>1). First, we define h:[0,1]R by the correspondence rule h(t)=rtlnt for t>0 and h(0)=0. The function h is concave and has a maximum at (1/e,r/e). Thus, we deduce that

rblnb<1,for all b[0,1] and r[0,e].
(2.1)

Secondary, by the Napier inequality [32]

0<b<a 1 a < ln a ln b a b < 1 b .
(2.2)

From (2.1) and (2.2) we have

rblnb1 1 a < ln a ln b a b ,

which implies γ>g(s). The proof of this case is completed by application of Proposition 2.1(ii).

Hence, by (a2), (b2), and (c2) we conclude that Theorem 1.1 is valid.

2.3 Proof of Theorem 1.2

The proof of this theorem is again developed by application of Proposition 2.1. Firstly, we recall the notation of [13]:

R + 3 = { ( a , b , c ) R 3 / a > 0 , b > 0  and  c > 0 } , E 1 = { ( a , b , c ) R + 3 / a = b = c  or  a = b c  or  a b = c } , E a + = { ( a , b , c ) R + 3 / a 1  and  a > max { b , c } } , E a = { ( a , b , c ) R + 3 / 1 > a > max { b , c } } , E b + = { ( a , b , c ) R + 3 / b 1  and  b > max { a , c } } , E b = { ( a , b , c ) R + 3 / 1 > b > max { a , c } } , E c + = { ( a , b , c ) R + 3 / c 1  and  c > max { a , b } } and E c = { ( a , b , c ) R + 3 / 1 > c > max { a , b } } .

The family { E 1 , E a + , E a , E b + , E b , E c + , E c } is a set partition of R + 3 . Now, with this notation, we subdivide the proof in three parts:

(a3) Case (a,b,c) E 1 . This special case is a direct consequence of Theorem 1.1.

(b3) Case (a,b,c) E a + E b + E c + . If (a,b,c) E a + , we apply Theorem 1.1 and Proposition 2.1 as follows. We select t= a r b , γ= c r b and s=a/b, the monotonic behavior and properties of function f, defined on Proposition 2.1, imply that

a r a + c r b > a r b + c r a ,
(2.3)

since t>γ, t>1 and s>1. Indeed, the corresponding proof of (2.3) needs the distinction of two cases: c1 and c<1. If c1, then γ>1 and γ]g(s),[, so f is strictly increasing and t>γ implies (2.3). For c<1, we note that γ<1 and γ s +γ0 since s>1 and 1]g(s),[, then the assumption t>1 implies that f(t)>f(1)= γ s +γ0=f(γ) and (2.3) is again true for this subcase. Moreover, for (a,b,c) E a + R + 3 , by Theorem 1.1, we recall that the inequality

c r c + b r b > b r c + c r b
(2.4)

holds true for all r[0,e]. Adding (2.3) and (2.4) we deduce (1.2).

The proof for (a,b,c) E b + E c + is similar to the case (a,b,c) E a + and we omit the details. However, we comment that for (a,b,c) E b + we choose t= b r c , γ= c 2 c , and s=b/c; and for (a,b,c) E c + we select t= c r a , γ= b r a , and s=c/a.

(c3) Case (a,b,c) E a E b E c . Without loss of generality, we assume that (a,b,c) E a is such that c<b<a, since the proof for b<c<a is similar. We note that Ω=[0,e]×[0,1] can be partitioned in the two sets

Ω 1 = { ( r , c ) Ω : c [ ( r 1 ) r 1 , 1 ] } and Ω 2 = { ( r , c ) Ω : c [ 0 , ( r 1 ) r 1 ] } .

Now, we continue the proof by distinguish the following two subcases: (r,c) Ω 1 and (r,c) Ω 2 .

For the subcase (r,c) Ω 1 , we apply the function f given on Proposition 2.1 with t= b r c , γ= c r c , and s=a/c to prove

b r a + c r c > b r c + c r a for 0<c<b<a<1 and (r,c) Ω 1 .
(2.5)

Indeed, we firstly note that the function m:[c,1]R defined as follows: m(z)=z c r z c r c + 1 has the following properties:

(m a ) m(c)=0;

(m b ) m(1)= c r (1 c r c + 1 r )0 for all (r,c) Ω 1 since c>(r1)/r; and

(m c ) m has a maximum at z max =1/rlnc, since the first and second derivatives of m are given by m (z)= c r z (1+rzlnc) and m (z)= c r z (2r+rzlnc)lnc and naturally m ( z max )=0 and m ( z max )<0.

Moreover, we notice that z max c is equivalent to 1>rclnc, which is true for r[0,e] and c[0,1]; see the proof of (2.1). Then, by (m a )-(m c ), it follows that m(z)0, for all z[c,1]. In particular, for z=a, we have

a c r a > c r c + 1 ,for a[c,1][0,1] and r[0,e].
(2.6)

Now, from (2.6), we note that

a c r a > c r c + 1 c r ( a c ) > c a r c ln c > c ln ( c / a ) a c c r c > e c ln ( a / c ) a c γ > g ( s ) ,
(2.7)

which implies (2.5) by application of Proposition 2.1(ii), since t>γ>g(s) and f is increasing on ]g(s),[.

For the subcase (r,c) Ω 2 , we apply the function f given on Proposition 2.1 with t= b r c , γ= c r c , and s=a/c to prove

b r a + c r c > b r c + c r a for 0<c<b<a<1 and (r,c) Ω 2 .
(2.8)

We note that the inequality c r c > c r 1 holds true for all (r,c) Ω 2 . Now, in order to deduce that γ>g(s) it is sufficient to prove that c r 1 >g(s). Indeed, the function q:[c,1]R defined as q(z)= c ( 1 r ) z z c c c + c ( 1 r ) has the following properties:

(q a ) q(c)=0;

(q b ) q(1)= c 1 r (1 c c + ( c 1 ) ( 1 r ) )0 for all (r,c) Ω 2 , since c[0,(r1)/r]; and

(q c ) q is increasing in [c,1].

Then we deduce that q(z)0 for all z[c,1]. In particular, for z=a[c,1], we deduce that c ( 1 r ) a a c c c + c ( 1 r ) 0, which implies the following sequence of implications:

c ( 1 r ) a a c > c c + c ( 1 r ) c ( 1 r ) a c c ( 1 r ) > c c a c c 1 r >g(s).

Thus (2.8) holds true.

From (2.5) and (2.8), we deduce that

b r a + c r c > b r c + c r a for 0<c<b<a<1 and r[0,e].
(2.9)

Hence, to complete the proof for 0<c<b<a<1, we add the inequality (2.9) with a r a + b r b > a r b + b r a for r[0,e], which is true by Theorem 1.1.

For (a,b,c) E b E c we can follow line by line the proof of (a,b,c) E a . However, we can obtain a direct proof by applying the result obtained for (a,b,c) E a by interchanging the role of variables. For instance, if (a,b,c) E b then (b,a,c) E a , which implies (1.2).

Hence, by (a3), (b3), and (c3) we have the complete proof of Theorem 1.2.

2.4 Proof of Theorem 1.3

Given b]0,1], we define the function H:]0,1]R as follows:

H(x)=2 x r x b r b b r x x r b .

Then we prove that H(x)>0 for all x]0,1], which naturally implies the inequality 2 a r a b r b a r b + b r a for x=a. Indeed, we prove that the function H has a global minimum at x=b. The fact that in x=b there is a local minimum of H follows by noticing that H (b)=0 and H (b)>0, since

H ( x ) = r [ x r x b r b ( ln x + 1 ) b r x ln b b x r b 1 ] and H ( x ) = r [ x r x b r b { r ( ln x + 1 ) 2 + x 1 } r b r x ( ln b ) 2 b ( r b 1 ) x r b 2 ] .

Meanwhile, the property that b is a global minimum of H can be proved by rewriting H as the difference of two functions and by analyzing the sign of H using some properties of this new functions. Indeed, to be more specific, we note that H (x)=r[K(x)Q(x)] for all x]0,1], where the functions K and Q are defined as follows:

K(x)= x r x b r b (lnx+1)andQ(x)= b r x lnb+b x r b 1 .

The functions K and Q have the following properties:

(K1) K is strictly increasing on ]0,1], since K (x)= x r x b r b {r ( ln x + 1 ) 2 + x 1 }>0, for all x]0,1].

(K2) K(x) when x 0 + , K(1/e)=0, and K(1)= b r b .

(Q1) The derivative of Q is given by Q (x)=r b r x ( ln b ) 2 +b(rb1) x r b 2 , for all x]0,1]. Then, in order to analyze the sign of Q , we introduce the set Λ=]0,1]×]0,e] and a partition { Λ 1 , Λ 2 , Λ 3 } of Λ, where

Λ 1 = { ( b , r ) Λ : Q ( x ) > 0  for all  x ] 0 , 1 ] } , Λ 2 = { ( b , r ) Λ : Q ( x ) < 0  for all  x ] 0 , 1 ] } , Λ 3 = { ( b , r ) Λ : ! c ] 0 , 1 ]  such that  Q  has a minimum at  x = c } .

We note that the sets Λ i , i=1,2,3, are not empty since for instance ]0,1[×[1/b,e] Λ 1 for all b]0,1], {1}×]0,1[ Λ 2 and ]0,1[×{1} Λ 3 . Moreover, we note that rb>1 implies that (b,r) Λ 1 and naturally Λ 2 Λ 3 is a subset of ]0,1]×]0,1/b[. The uniqueness of c can be deduced by noticing that the solution of Q (x)=0 is equivalent to the intersection of the following two monotone functions: S(x)=r b r x ( ln b ) 2 and J(x)=b(1rb) x r b 2 .

(Q2) Q(x)lnb when x 0 + , and Q(1)= b r lnb+b.

From (K1) and (Q1) we deduce the uniqueness of b]0,1] such that Q(b)=K(b) or equivalently H (b)=0. Now, from (K2) and (Q2), we note that Q( 0 + )>K( 0 + ) for all (r,b)Λ since K( 0 + )=. Then H (x)<0 for all x]0,b[. Additionally, from (K2) and (Q2), we observe that Q(1)<K(1). This fact is a consequence of the fact that the function F(w,r)= w r w w r ln(w)w is strictly decreasing in r, since F r (w,r)=ln(w)((r/2) w r w w r ln(w))<0. Consequently, for r<e we have F(w,r)>F(w,e)= w e w w e ln(w)w>0 for all w]0,1]. Hence, for w=b we get F(b,r)>0 or Q(1)<K(1), which implies that H (x)>0 for all x]b,1]. Thus, b is a global minimum of H. Therefore, H(x)H(b)=0 for all x]0,1] and in particular for x=a.

2.5 Proof of Theorem 1.4

The proof follows by the fact that the function P: ] 0 , 1 ] n 1 R is defined by the following correspondence rule:

P( z 1 ,, z n 1 )=n x n x n i = 1 n z i z i ( x n j = 1 n 1 z j ) x n i = 1 n 1 ( x n j = 1 n 1 z j ) x i , x n ]0,1],

and it has a global minimum at ( z 1 ,, z n 1 )=( x n ,, x n ). Indeed, for simplicity of notation we develop the details of the proof for n=3 and with ( x 1 , x 2 , x 3 )=(a,b,c). Note that, in this case for an arbitrary c]0,1], the function P: ] 0 , 1 ] 2 R has the following form:

P(x,y)=3 x x y y c c ( x y c ) x ( x y c ) y ( x y c ) c .

Then we have

P x ( x , y ) = 3 x x y y c c ( ln ( x ) + 1 ) ( ln ( x y c ) + 1 ) ( x y c ) x y x ( x y c ) y c x ( x y c ) c , P y ( x , y ) = 3 x x y y c c ( ln ( y ) + 1 ) x y ( x y c ) x ( ln ( x y c ) + 1 ) ( x y c ) y c y ( x y c ) c , P x x ( x , y ) = 3 x x y y c c [ 1 x + ( ln ( x ) + 1 ) 2 ] [ 1 x + ( ln ( x y c ) + 1 ) 2 ] ( x y c ) x [ y 2 y x 2 ] ( x y c ) y [ c 2 c x 2 ] ( x y c ) c , P y y ( x , y ) = 3 x x y y c c [ 1 y + ( ln ( y ) + 1 ) 2 ] [ x 2 x y 2 ] ( x y c ) x [ 1 y + ( ln ( x y c ) + 1 ) 2 ] ( x y c ) y [ c 2 c y 2 ] ( x y c ) c , P x y ( x , y ) = P y x ( x , y ) = 3 x x y y c c ( ln ( y ) + 1 ) ( ln ( x ) + 1 ) [ x ( ln ( x y c ) + 1 ) + 1 y ] ( x y c ) x [ y ( ln ( x y c ) + 1 ) + 1 x ] ( x y c ) y [ c 2 x y ] ( x y c ) c .

An evaluation at (c,c) implies that

P x ( c , c ) = P y ( c , c ) = 0 , P x x ( c , c ) = P y y ( c , c ) = c 3 c 1 ( 6 c ( ln ( c ) ) 2 + 4 ) , P x y ( c , c ) = P y x ( c , c ) = c 3 c 1 ( 3 c ( ln ( c ) ) 2 2 ) .

Now, defining P 1 (w)=6w ( ln ( w ) ) 2 +4 and P 2 (w)=27 w 2 ( ln w ) 4 24w ( ln w ) 2 +4, we observe that P x x (c,c)= c 3 c 1 P 1 (c) and P x x (c,c) P y y (c,c) P x y (c,c) P x y (c,c)= c 2 ( 3 c 1 ) P 2 (c). Then the Hessian matrix associated to P at (c,c) is positive semidefinite since both functions, P 1 and P 2 , are positive on ]0,1] or equivalently the function P has a local minimum at (c,c). Now, we deduce that (c,c) is the global minimum since we can prove that (c,c) is the unique solution of ( P x , P y )=(0,0). Indeed, assuming that there is (x,y) with xyc such that P x (x,y)= P y (x,y)=0, we can deduce a contradiction. Note that

0 = | P x ( x , y ) P y ( x , y ) | | min { 3 x x y y c c , ( x y c ) x , ( x y c ) y , ( x y c ) c } | | ln ( x y ) y x c x + x y + c y | | min { 3 x x y y c c , ( x y c ) x , ( x y c ) y , ( x y c ) c } | | 1 x + x + y x y + c x y | | x y | ,

since the inequality ln(r)>(r1)/r holds for all r>0 and r1 (see for instance [27]). Then x=y, which is a contradiction with the assumption that xy. Thus, we see that (c,c) is a global minimum of the function P or equivalently P(x,y)P(c,c)=0 for all (x,y) ] 0 , 1 ] 2 , which implies the desired inequality for (x,y)=(a,b).

3 Additional remarks on possible generalizations

In this section we present the possible extensions of Theorems 1.1, 1.2, and 1.3 to a sequence of positive real numbers. We note that the natural generalizations of (1.2) and (1.3) are given by

i = 1 n x i r x i x n r x 1 + i = 1 n 1 x i r x i + 1 ,( x 1 ,, x n ) R + n ,r[0,e],
(3.1)
n i = 1 n x i r x i n x n r x 1 + i = 1 n 1 x i r x i + 1 ,( x 1 ,, x n ) ] 0 , 1 ] n ,
(3.2)

respectively. We present a partial proof of (3.1) (see Lemma 3.1, below) and leave as a conjecture the proof of (3.2).

Lemma 3.1 The inequality given in (3.1) holds for all r[0,e], if we restrict ( x 1 ,, x n ) to the hypercube [ 0 , 1 ] n .

Proof Before we start the proof, we notice that the function ϒ(x,y)= x a / b x y a / b +y defined from R + 2 R and for a>b is concave and ϒ(0,0)=ϒ(1,0)=ϒ(0,1)=ϒ(1,1)=0. Then ϒ(x,y)0 for all (x,y)[0,1]×[0,1]. Similarly, the function ϒ s (w,z)=ϒ(z,w) for a<b is concave and ϒ s (w,z)0 for all (w,z)[0,1]×[0,1]. Now, we proceed by induction on n. Let us assume that the theorem is valid for a sequence of positive numbers ( x 1 ,, x k ) for all k<n. We note that

i = 1 n x i r x i x n r x 1 i = 1 n 1 x i r x i + 1 = [ i = 1 n 1 x i r x i x n 1 r x 1 i = 1 n 2 x i r x i + 1 ] + [ x n r x n + x n 1 r x n 1 x n r x n 1 x n 1 r x n ] + [ x n r x n 1 x n r x 1 x n 1 r x n 1 + x n 1 r x 1 ] : = K 1 + K 2 + K 3 .
(3.3)

The terms K 1 and K 2 are positive by the inductive hypothesis. Meanwhile, the term K 3 is positive by the concavity of the functions ϒ and ϒ s . Note that a= x n 1 and b= x n , and K 3 =ϒ( x n r x 1 , x n 1 r x 1 ) or K 3 = ϒ s ( x n 1 r x 1 , x n r x 1 ), depending if x n 1 > x 1 or x n 1 < x 1 , respectively. Then, by (3.3), it follows that the lemma is valid. □

Conjecture 3.1 Let nN and n>4. Then the inequality (3.1) holds for all ( x 1 ,, x n ) R + n and r[0,e].

Conjecture 3.2 Let nN and n3. Then the inequality (3.2) holds for all r[0,e].

Conjecture 3.3 Let nN and x i ] 0 , 1 ] n . Then the inequality

n i = 1 n x i r x i i = 1 n ( j = 1 n x j ) r x i

holds for all r[0,e].

References

  1. Dahmani A, Karima Belaide K: Exponential inequalities in calibration problems with gaussians errors. Commun. Stat., Theory Methods 2013,42(19):3596-3607. 10.1080/03610926.2011.635257

    Article  MathSciNet  MATH  Google Scholar 

  2. Hou Q, Lin Z, Dusing RW, Gajewski BJ, Mccallum RW: A bayesian hierarchical assessment of gastric emptying with the linear, power exponential and modified power exponential models. Neurogastroenterology & Motility 2010,22(12):1308-1317. 10.1111/j.1365-2982.2010.01572.x

    Article  Google Scholar 

  3. Floyd BNI, Camilleri M, Andresen V, Esfandyari T, Busciglio I, Zinsmeister AR: Comparison of mathematical methods for calculating colonic compliance in humans: power exponential, computer-based and manual linear interpolation models. Neurogastroenterology & Motility 2008,20(4):330-335. 10.1111/j.1365-2982.2007.01024.x

    Article  Google Scholar 

  4. Park J-S, Baek J: Efficient computation of maximum likelihood estimators in a spatial linear model with power exponential covariogram. Comput. Geosci. 2001,27(1):1-7. 10.1016/S0098-3004(00)00016-9

    MathSciNet  Article  Google Scholar 

  5. Bruno AD: Power-exponential expansions of solutions to an ordinary differential equation. Dokl. Math. 2012,85(3):336-340. 10.1134/S106456241203009X

    MathSciNet  Article  MATH  Google Scholar 

  6. Fan X, Grama I, Liu Q: Large deviation exponential inequalities for supermartingales. Electron. Commun. Probab. 2012,17(59):1-8.

    MathSciNet  MATH  Google Scholar 

  7. Miyagi M, Nishizawa Y: A short proof of an open inequality with power-exponential functions. Aust. J. Math. Anal. Appl. 2014. Article ID 6, 11(1): Article ID 6

    MATH  Google Scholar 

  8. Miyagi M, Nishizawa Y: Proof of an open inequality with double power-exponential functions. J. Inequal. Appl. 2013. Article ID 468, 2013: Article ID 468

    Google Scholar 

  9. Li Y: Solutions of two conjectures on inequalities with power-exponential functions. RGMIA Res. Rep. Collect. 2009. Article ID 7, 12(4): Article ID 7

    Google Scholar 

  10. Hisasue M: Solution of inequalities with power-exponential functions by Cîrtoaje. Aust. J. Math. Anal. Appl. 2012. Article ID 4, 9(2): Article ID 4

    MATH  Google Scholar 

  11. Cîrtoaje V: Proofs of three open inequalities with power-exponential functions. J. Nonlinear Sci. Appl. 2011,4(2):130-137.

    MathSciNet  MATH  Google Scholar 

  12. Cîrtoaje V: On some inequalities with power-exponential functions. J. Inequal. Pure Appl. Math. 2009. Article ID 21, 10(1): Article ID 21

    MATH  Google Scholar 

  13. Coronel A, Huancas F:On the inequality a 2 a + b 2 b + c 2 c a 2 b + b 2 c + c 2 a . Aust. J. Math. Anal. Appl. 2012. Article ID 3, 9(1): Article ID 3

    MATH  Google Scholar 

  14. Manyama S: Solution of one conjecture on inequalities with power-exponential functions. Aust. J. Math. Anal. Appl. 2010. Article ID 1, 7(2): Article ID 1

    MATH  Google Scholar 

  15. Matejíčka L: On an open problem posed in the paper ‘Inequalities of power-exponential functions’. J. Inequal. Pure Appl. Math. 2008. Article ID 75, 9(3): Article ID 75

    MATH  Google Scholar 

  16. Qi F, Debnath L: Inequalities for power-exponential functions. J. Inequal. Pure Appl. Math. 2000. Article ID 15, 1(2): Article ID 15

    MATH  Google Scholar 

  17. Qi F, Xu S-L:The function ( b x a x )/x: inequalities and properties. Proc. Am. Math. Soc. 1998,126(11):3355-3359. 10.1090/S0002-9939-98-04442-6

    MathSciNet  Article  MATH  Google Scholar 

  18. Cerone P, Dragomir SS: Mathematical Inequalities. CRC Press, Boca Raton; 2011.

    MATH  Google Scholar 

  19. Wright EM:Solution of the equation z e z =a. Proc. R. Soc. Edinb., Sect. A 1959, 65: 193-203.

    MATH  Google Scholar 

  20. Wright EM:Solution of the equation z e z =a. Bull. Am. Math. Soc. 1959, 65: 89-93. 10.1090/S0002-9904-1959-10290-1

    Article  MATH  Google Scholar 

  21. Wright EM:Solution of the equation (pz+q) e z =rz+s. Bull. Am. Math. Soc. 1960, 66: 277-281. 10.1090/S0002-9904-1960-10469-7

    Article  MATH  Google Scholar 

  22. Lambert, JH: Observations variae in mathesin puram. Acta Helvitica, Physico-Mathematico-Anatomico-Botanico-Medica 63, 128-168 (1758)

  23. Euler, J: De serie lambertina plurimisque eius insignibus proprietatibus. Acta Acad. Scient. Petropol. 2, 29-51 (1783)

  24. Pólya G, Szegő G: Aufgaben und Lehrsätze aus der Analysis. Band II: Funktionentheorie, Nullstellen, Polynome Determinanten, Zahlentheorie. Springer, Berlin; 1971.

    Book  MATH  Google Scholar 

  25. Corless RM, Gonnet GH, Hare DEG, Jeffrey DJ, Knuth DE: On the Lambert W function. Adv. Comput. Math. 1996,5(4):329-359.

    MathSciNet  Article  MATH  Google Scholar 

  26. Hoorfar A, Hassani M: Inequalities on the Lambert W function and hyperpower function. J. Inequal. Pure Appl. Math. 2008. Article ID 51, 9(2): Article ID 51

    MATH  Google Scholar 

  27. Bullen PS Pitman Monographs and Surveys in Pure and Applied Mathematics 97. In A Dictionary of Inequalities. Longman, Harlow; 1998.

    Google Scholar 

  28. Luo J, Wen JJ:A power-mean discriminance of comparing a b and b a . In Research Inequalities. Edited by: Yand X-Z. People’s Press of Tibet, The People’s Republic of China; 2000:83-88.

    Google Scholar 

  29. Cîrtoaje V: Proofs of three open inequalities with power-exponential functions. J. Nonlinear Sci. Appl. 2011,4(2):130-137.

    MathSciNet  MATH  Google Scholar 

  30. zeikii, A, Cîrtoaje, V, Berndt, B:. Mathlinks Forum. http://www.mathlinks.ro/Forum/viewtopic.php?t=118722 (2006). Accessed November 2006

  31. Matejíčka L: Solution of one conjecture on inequalities with power-exponential functions. J. Inequal. Pure Appl. Math. 2009,10(3):1-5.

    MathSciNet  MATH  Google Scholar 

  32. Nelsen RB: Napier’s inequality (two proofs). Coll. Math. J. 1993. Article ID 165, 24(2): Article ID 165

    Google Scholar 

Download references

Acknowledgements

We acknowledge the support of ‘Univesidad del Bío-Bío’ (Chile) through the research projects 124109 3/R, 104709 01 F/E, and 121909 GI/C.

Author information

Authors and Affiliations

Authors

Corresponding author

Correspondence to Aníbal Coronel.

Additional information

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

Rights and permissions

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (https://creativecommons.org/licenses/by/4.0), which permits use, duplication, adaptation, distribution, and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Reprints and Permissions

About this article

Verify currency and authenticity via CrossMark

Cite this article

Coronel, A., Huancas, F. The proof of three power-exponential inequalities. J Inequal Appl 2014, 509 (2014). https://doi.org/10.1186/1029-242X-2014-509

Download citation

  • Received:

  • Accepted:

  • Published:

  • DOI: https://doi.org/10.1186/1029-242X-2014-509

Keywords

  • power inequalities
  • exponential inequalities
  • power-exponential inequalities