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An artificial proof of a geometric inequality in a triangle

Abstract

In this paper, the authors give an artificial proof of a geometric inequality relating to the medians and the exradius in a triangle by making use of certain analytical techniques for systems of nonlinear algebraic equations.

MSC:51M16, 52A40.

1 Introduction and main results

For a given ABC, let a, b and c denote the side-lengths facing the angles A, B and C, respectively. Also, let m a , m b and m c denote the corresponding medians, r a , r b and r c the corresponding exradii, s= 1 2 (a+b+c) the semi-perimeter, Δ the area. In addition, we let

m 1 = 1 2 ( b + c ) 2 a 2 = s ( s a ) , m 2 = 1 2 2 a 2 + 1 4 ( b + c ) 2 ,

and

r 1 = a s ( s a ) 2 ( s a ) .

Throughout this paper, we will customarily use the cyclic sum symbols as follows:

f(a)=f(a)+f(b)+f(c)

and

f(b,c)=f(a,b)+f(b,c)+f(c,a).

In 2003, Liu [1] found the following interesting geometric inequality relating to the medians and the exradius in a triangle with the computer software BOTTEMA invented by Yang [25], and Liu thought this inequality cannot be proved by a human.

Theorem 1.1 In ABC, the best constant k for the following inequality

( r b r c ) 2 k ( m b m c ) 2
(1.1)

is the real root on the interval (3,4) of the following equation

6,561 k 4 14,256 k 3 18,080 k 2 25,344k+20,736=0.
(1.2)

Furthermore, the constant k has its numerical approximation given by 3.2817755127.

In this paper, the authors give an artificial proof of Theorem 1.1.

2 Preliminary results

In order to prove Theorem 1.1, we require the following results.

Lemma 2.1 In ABC, if abc, then

r a 2 + r b 2 + r c 2 ( r 1 2 + 2 m 1 2 ) 3 s ( s a ) ( b c ) 2 4 ( s b ) ( s c ) .
(2.1)

Proof From a=(sb)+(sc) and the formulas of the exradius r a = Δ s a = s ( s a ) ( s b ) ( s c ) s a , etc., we get

r a 2 + r b 2 + r c 2 ( r 1 2 + 2 m 1 2 ) = [ 1 ( s a ) 2 + 1 ( s b ) 2 + 1 ( s c ) 2 ] s ( s a ) ( s b ) ( s c ) a 2 s ( s a ) 4 ( s a ) 2 2 s ( s a ) = 1 4 s ( s a ) [ 4 ( s b ) ( s c ) ( s a ) 2 + 4 ( s b ) ( s c ) ( s b ) 2 + 4 ( s b ) ( s c ) ( s c ) 2 a 2 ( s a ) 2 8 ] = 1 4 s ( s a ) [ 4 ( s b ) ( s c ) a 2 ( s a ) 2 + 4 ( s c s b + s b s c 2 ) ] = 1 4 s ( s a ) [ ( b c ) 2 ( s a ) 2 + 4 ( b c ) 2 ( s b ) ( s c ) ] = 1 4 s ( s a ) ( b c ) 2 [ 4 ( s b ) ( s c ) 1 ( s a ) 2 ] .
(2.2)

For abc, we have

sasbsc>0,

then

0< 1 s a 1 s b 1 s c ,

hence

1 ( s b ) ( s c ) 1 ( s a ) 2 >0.
(2.3)

Inequality (2.1) follows from inequalities (2.2)-(2.3) immediately. □

Lemma 2.2 In ABC, we have

( m b + m 2 )( m c + m 2 )4s ( s b ) ( s c )
(2.4)

and

a ( m b + m c ) 2 8s(sb)(sc) 3 s ( s b ) ( s c ) ( b c ) 2 a .
(2.5)

Proof of inequality (2.4) From

m 2 2 1 2 as= 1 4 ( a b + c 2 ) 2 0,

we immediately obtain

m 2 1 2 a s .
(2.6)

In view of the AM-GM inequality, we get

a 2 = ( s b ) + ( s c ) 2 ( s b ) ( s c ) .
(2.7)

By the power mean inequality, we have

a 2 = ( s b ) + ( s c ) 2 s b + s c 2 .
(2.8)

By the well-known inequalities m b s ( s b ) and m c s ( s c ) , together with inequalities (2.6)-(2.8), we obtain

( m b + m 2 ) ( m c + m 2 ) ( s ( s b ) + 1 2 a s ) ( s ( s c ) + 1 2 a s ) = s ( s b + 1 2 a ) ( s c + 1 2 a ) = s [ 1 2 a + 1 2 a ( s b + s c ) + ( s b ) ( s c ) ] s [ 1 2 ( s b + s c ) 2 + 2 ( s b ) ( s c ) ] = s [ 1 2 a + 3 ( s b ) ( s c ) ] 4 s ( s b ) ( s c ) .

The proof of inequality (2.4) is thus complete. □

Proof of inequality (2.5) According to the well-known inequalities m b s ( s b ) , m c s ( s c ) and inequality (2.7), we have

a ( m b + m c ) 2 8 s ( s b ) ( s c ) = [ a 2 ( s b ) ( s c ) ] ( m b + m c ) 2 + 2 ( s b ) ( s c ) [ ( m b + m c ) 2 4 s ( s b ) ( s c ) ] [ a 2 ( s b ) ( s c ) ] 4 m b m c + 2 ( s b ) ( s c ) [ ( s ( s b ) + s ( s c ) ) 2 4 s ( s b ) ( s c ) ] 4 s [ a 2 ( s b ) ( s c ) ] ( s b ) ( s c ) + 2 ( s b ) ( s c ) [ a 2 ( s b ) ( s c ) ] = 6 s ( s b ) ( s c ) [ a 2 ( s b ) ( s c ) ] = 6 s ( s b ) ( s c ) ( b c ) 2 a + 2 ( s b ) ( s c ) 3 s ( s b ) ( s c ) ( b c ) 2 a .
(2.9)

Hence, we complete the proof of inequality (2.5). □

Lemma 2.3 In ABC, we have

m b m c m 2 2 .
(2.10)

Proof From the formulas of the medians, we have

m b m c m 2 2 = m b 2 m c 2 m 2 4 m b m c + m 2 2 = 1 16 ( 2 c 2 + 2 a 2 b 2 ) ( 2 a 2 + 2 b 2 c 2 ) 1 16 ( 2 a 2 + 1 4 ( b + c ) 2 ) m b m c + m 2 2 = { 16 [ a 2 ( b + c ) 2 ] ( 17 b 2 + 17 c 2 + 38 b c ) } ( b c ) 2 256 ( m b m c + m 2 2 ) 0 .

Therefore, inequality (2.10) holds true. □

Lemma 2.4 In ABC, if abc, then

m b + m c m a + m 1 + 1 4 ( m 1 m 2 + m b + m 1 m 2 + m c ) 9 ( b + c ) 2 8 ( m b + m c ) 2 m 2 m 1 + m 1 4 m 2 9 a ( b + c ) 2 64 s ( s b ) ( s c ) .
(2.11)

Proof It is obvious that m b >c b 2 and m c >b c 2 , then we have m b + m c > 1 2 (b+c), thus

( m b m c ) 2 = ( m b 2 m c 2 ) 2 ( m b + m c ) 2 = 9 ( b + c ) 2 ( b c ) 2 16 ( m b + m c ) 2 9 4 ( b c ) 2 .
(2.12)

For abc, we have that

m a { m 1 m b m 2 m c .
(2.13)

By Lemma 2.3 and inequalities (2.12)-(2.13), we have

m b + m c m a + m 1 + 1 4 ( m 1 m 2 + m b + m 1 m 2 + m c ) m 2 m 1 m 1 4 m 2 = m b + m c 2 m 2 m a + m 1 + m 2 ( m 1 m a ) m 1 ( m a + m 1 ) + m 1 ( m 2 2 m b m c ) 4 m 2 ( m 2 + m b ) ( m 2 + m c ) ( m b + m c ) 2 4 m 2 2 ( m a + m 1 ) ( m b + m c + 2 m 2 ) + m 2 ( m 1 2 m a 2 ) m 1 ( m a + m 1 ) 2 = 2 ( m b 2 + m c 2 ) ( m b m c ) 2 4 m 2 2 ( m a + m 1 ) ( m b + m c + 2 m 2 ) + m 2 ( m 1 2 m a 2 ) m 1 ( m a + m 1 ) 2 = 1 4 ( b c ) 2 ( m b m c ) 2 ( m a + m 1 ) ( m b + m c + 2 m 2 ) m 2 ( b c ) 2 4 m 1 ( m a + m 1 ) 2 1 4 ( b c ) 2 9 4 ( b c ) 2 ( m a + m 1 ) ( m b + m c + 2 m 2 ) ( b c ) 2 4 ( m a + m 1 ) 2 = 2 ( b c ) 2 ( m a + m 1 ) ( m b + m c + 2 m 2 ) ( b c ) 2 4 ( m a + m 1 ) 2 2 ( b c ) 2 ( m a + m 1 ) ( m b + m c ) ( b c ) 2 4 ( m a + m 1 ) 2 = 2 ( b c ) 2 ( m b + m c ) 2 ( b c ) 2 4 ( m b + m c ) 2 9 ( b c ) 2 4 ( m b + m c ) 2 .
(2.14)

By inequality (2.5), (2.7) and abc, we obtain that

9 a ( b + c ) 2 64 s ( s b ) ( s c ) 9 ( b + c ) 2 8 ( m b + m c ) 2 = 9 ( b + c ) 2 [ a ( m b + m c ) 2 8 s ( s b ) ( s c ) ] 64 s ( s b ) ( s c ) ( m b + m c ) 2 9 ( b + c ) 2 64 s ( s b ) ( s c ) ( m b + m c ) 2 3 s ( s b ) ( s c ) ( b c ) 2 a = 27 ( b + c ) 2 ( b c ) 2 64 a ( s b ) ( s c ) ( m b + m c ) 2 27 ( b + c ) 2 ( b c ) 2 32 a 2 ( m b + m c ) 2 27 ( b c ) 2 8 ( m b + m c ) 2 .
(2.15)

By inequalities (2.14)-(2.15), we have

[ m b + m c m a + m 1 + 1 4 ( m 1 m 2 + m b + m 1 m 2 + m c ) 9 ( b + c ) 2 8 ( m b + m c ) 2 ] [ m 2 m 1 + m 1 4 m 2 9 a ( b + c ) 2 64 s ( s b ) ( s c ) ] = [ m b + m c m a + m 1 + 1 4 ( m 1 m 2 + m b + m 1 m 2 + m c ) m 2 m 1 m 1 4 m 2 ] + [ 9 a ( b + c ) 2 64 s ( s b ) ( s c ) 9 ( b + c ) 2 8 ( m b + m c ) 2 ] 9 ( b c ) 2 4 ( m b + m c ) 2 + 27 ( b c ) 2 8 ( m b + m c ) 2 = 9 ( b c ) 2 8 ( m b + m c ) 2 0 .
(2.16)

Inequality (2.11) follows from inequality (2.16) immediately. □

Lemma 2.5 In ABC, if abc, then

m 2 m 1 + m 1 4 m 2 + 3 ( b + c ) 2 16 a 2 2
(2.17)

and

m 1 + 3 a s 3 .
(2.18)

Proof Without loss of generality, we can take b+c=2 and a=x, for abc, we have 0<x1.

  1. (i)

    First, we prove inequality (2.17).

    m 2 m 1 + m 1 4 m 2 + 3 ( b + c ) 2 16 a 2 2 = 1 + 2 x 2 4 x 2 + 1 4 4 x 2 1 + 2 x 2 + 3 4 x 2 2 = 8 + 7 x 2 4 ( 4 x 2 ) ( 1 + 2 x 2 ) + 3 ( 1 x 2 ) 4 x 2 5 4 8 + 7 x 2 4 ( 4 x 2 ) + ( 1 + 2 x 2 ) 2 + 3 ( 1 x 2 ) 4 x 2 5 4 = 8 + 7 x 2 2 ( 5 + x 2 ) + 3 ( 1 x 2 ) 4 x 2 5 4 = 9 ( x 2 1 ) 4 ( 5 + x 2 ) + 3 ( 1 x 2 ) 4 x 2 3 ( x 2 1 ) 8 + 3 ( 1 x 2 ) 4 = 3 ( 1 x 2 ) 8 0 .
    (2.19)

Inequality (2.19) terminates the proof of inequality (2.17).

  1. (ii)

    Second, we prove inequality (2.18).

    m 1 + 3 a 3 s = 1 2 4 x 2 3 2 ( 2 x ) = 1 2 2 x ( 2 + x 3 ( 2 x ) ) = 2 2 x ( 1 x ) 2 + x + 3 ( 2 x ) 0 .
    (2.20)

Inequality (2.18) follows from inequality (2.20) immediately. □

Lemma 2.6 In ABC, if abc, then

m a m b + m b m c + m c m a 2 m 1 m 2 m 2 2 3 8 ( b c ) 2 3 s ( s a ) ( b c ) 2 16 ( s b ) ( s c ) .
(2.21)

Proof By the AM-GM inequality, the well-known inequalities m b s ( s b ) and m c s ( s c ) , we get

( m b + m c ) 2 4 m b m c 4s ( s b ) ( s c ) 6a ( s b ) ( s c ) 12(sb)(sc)

or

m b + m c 2 3 ( s b ) ( s c ) .
(2.22)

By inequalities (2.4), (2.10), (2.11), (2.17), (2.22), we obtain that

m a m b + m b m c + m c m a 2 m 1 m 2 m 2 2 = ( m b + m c ) ( m a 2 m 1 2 ) m a + m 1 + m 1 ( m b 2 m 2 2 ) m b + m 2 + m 1 ( m c 2 m 2 2 ) m c + m 2 ( m b 2 m c 2 ) 2 2 ( m b + m c ) 2 + 1 16 ( b c ) 2 = ( m b + m c ) ( b c ) 2 4 ( m a + m 1 ) + m 1 ( 5 b + 7 c ) ( c b ) 16 ( m b + m 2 ) + m 1 ( 7 b + 5 c ) ( b c ) 16 ( m c + m 2 ) 9 ( b + c ) 2 ( b c ) 2 32 ( m b + m c ) 2 + 1 16 ( b c ) 2 = ( m b + m c ) ( b c ) 2 4 ( m a + m 1 ) + m 1 ( b c ) 2 16 ( m b + m 2 ) + m 1 ( b c ) 2 16 ( m c + m 2 ) 9 m 1 ( b + c ) 2 ( b c ) 2 32 ( m b + m 2 ) ( m c + m 2 ) ( m b + m c ) 9 ( b + c ) 2 ( b c ) 2 32 ( m b + m c ) 2 + 1 16 ( b c ) 2 1 4 ( m 2 m 1 + m 1 4 m 2 9 a ( b + c ) 2 64 s ( s b ) ( s c ) m 1 ( b + c ) 2 64 3 s ( s b ) ( s c ) + 1 4 ) ( b c ) 2 = 1 4 ( m 2 m 1 + m 1 4 m 2 9 ( m 1 + 3 a ) ( b + c ) 2 64 3 s ( s b ) ( s c ) + 1 4 ) ( b c ) 2 1 4 ( m 2 m 1 + m 1 4 m 2 9 ( b + c ) 2 64 ( s b ) ( s c ) + 1 4 ) ( b c ) 2 = 1 4 ( m 2 m 1 + m 1 4 m 2 3 [ ( b + c ) 2 a 2 ] 16 ( s b ) ( s c ) + 3 [ ( b + c ) 2 4 a 2 ] 64 ( s b ) ( s c ) + 1 4 ) ( b c ) 2 1 4 ( m 2 m 1 + m 1 4 m 2 3 s ( s a ) 4 ( s b ) ( s c ) + 3 [ ( b + c ) 2 4 a 2 ] 16 a 2 + 1 4 ) ( b c ) 2 = 1 4 ( m 2 m 1 + m 1 4 m 2 + 3 ( b + c ) 2 16 a 2 3 s ( s a ) 4 ( s b ) ( s c ) 1 2 ) ( b c ) 2 1 4 ( 2 3 s ( s a ) 4 ( s b ) ( s c ) 1 2 ) ( b c ) 2 = 3 8 ( b c ) 2 3 s ( s a ) ( b c ) 2 16 ( s b ) ( s c ) .

The proof of Lemma 2.6 is thus completed. □

Lemma 2.7 In ABC, if inequality (1.1) holds, then k4.

Proof Let b=c=1 and a=x. For abc, we have x(0,1], then inequality (1.1) is equivalent to

2 ( x 4 x 2 2 ( 2 x ) 4 x 2 2 ) 2 2 k ( 4 x 2 2 2 x 2 + 1 2 ) 2 2 + x 2 x k 9 ( 1 + x ) 2 4 ( 4 x 2 + 2 x 2 + 1 ) 2 k 4 ( 2 + x ) ( 4 x 2 + 2 x 2 + 1 ) 2 9 ( 2 x ) ( 1 + x ) 2 .
(2.23)

Taking x=1 in inequality (2.23), we obtain that k4. □

Lemma 2.8 In ABC, if abc and 0<k4, then we have

( r b r c ) 2 k ( m b m c ) 2 2 ( r 1 m 1 ) 2 2k ( m 1 m 2 ) 2 .
(2.24)

Proof For

( r b r c ) 2 =2 r a 2 2 r b r c =2 r a 2 2 s 2

and

( m b m c ) 2 =2 m a 2 2 m b m c = 3 2 a 2 2 m b m c ,

hence, by Lemmas 2.1 and 2.6, we have

( r b r c ) 2 k ( m b m c ) 2 2 ( r 1 m 1 ) 2 + 2 k ( m 1 m 2 ) 2 = 2 [ r a 2 r 1 2 2 m 1 2 ] + 2 k [ m b m c 2 m 1 m 2 m 2 2 3 8 ( b c ) 2 ] 3 s ( s a ) ( b c ) 2 2 ( s b ) ( s c ) 3 k s ( s a ) ( b c ) 2 8 ( s b ) ( s c ) = 3 ( 4 k ) s ( s a ) ( b c ) 2 8 ( s b ) ( s c ) 0 .

The proof of Lemma 2.8 is complete. □

Lemma 2.9 (see [4, 6, 7])

Define

F(x)= a 0 x n + a 1 x n 1 ++ a n ,

and

G(x)= b 0 x m + b 1 x m 1 ++ b m .

If a 0 0 or b 0 0, then the polynomials F(x) and G(x) have a common root if and only if

where R(F,G) ((m+n)×(m+n) determinant) is Sylvester’s resultant of F(x) and G(x).

Lemma 2.10 (see [7, 8])

Given a polynomial f(x) with real coefficients

f(x)= a 0 x n + a 1 x n 1 ++ a n ,

if the number of the sign changes in the revised sign list of its discriminant sequence

{ D 1 ( f ) , D 2 ( f ) , , D n ( f ) }

is v, then the number of the pairs of distinct conjugate imaginary roots of f(x) equals v. Furthermore, if the number of non-vanishing members in the revised sign list is l, then the number of the distinct real roots of f(x) equals l2v.

3 The proof of Theorem 1.1

Proof If k0,we can easily find that inequality (1.1) holds. Hence, we only need to consider the case k>0, and by Lemma 2.7, we only need to consider the case 0<k4.

Now we determine the best constant k such that inequality (1.1) holds. Since inequality (1.1) is symmetrical with respect to the side-lengths a, b and c, there is no harm in supposing abc. Thus, by Lemma 2.8, we only need to determine the best constant k such that

2 ( r 1 m 1 ) 2 2k ( m 1 m 2 ) 2 0

or, equivalently, that

( a ( b + c ) 2 a 2 2 ( b + c a ) ( b + c ) 2 a 2 2 ) 2 k ( ( b + c ) 2 a 2 2 1 2 2 a 2 + 1 4 ( b + c ) 2 ) 2 0 .
(3.1)

Without loss of generality, we can assume that

a=tand b + c 2 =1(0<t1),

because inequality (3.1) is homogeneous with respect to a and b + c 2 . Thus, clearly, inequality (3.1) is equivalent to the following inequality:

( t 4 t 2 2 ( 2 t ) 4 t 2 2 ) 2 k ( 4 t 2 2 2 t 2 + 1 2 ) 2 0.
(3.2)

We consider the following two cases separately.

Case 1. When t=1, inequality (3.2) holds true for any kR:=(,+).

Case 2. When 0<t<1, inequality (3.2) is equivalent to the following inequality:

k 4 ( 2 + t ) ( 4 t 2 + 2 t 2 + 1 ) 2 9 ( 2 t ) ( 1 + t ) 2 .
(3.3)

Define the function

g(t):= 4 ( 2 + t ) ( 4 t 2 + 2 t 2 + 1 ) 2 9 ( 2 t ) ( 1 + t ) 2 ,x(0,1).

Calculating the derivative for g(t), we get

g (t)= 8 ( 4 t 2 + 2 t 2 + 1 ) 4 t 2 [ ( 2 t 3 + 5 t 2 + 10 t 2 ) ( 2 t ) 4 t 2 2 t 2 + 1 ] 9 ( 2 t ) 2 ( 1 + t ) 3 2 t 2 + 1 4 t 2 .

By setting g (t)=0, we obtain

4 t 2 [ ( 2 t 3 + 5 t 2 + 10 t 2 ) ( 2 t ) 4 t 2 2 t 2 + 1 ] =0.
(3.4)

It is easily observed that the equation 4 t 2 =0 has no real root on the interval (0,1). Hence, the roots of equation (3.4) are also solutions of the following equation:

( 2 t 3 + 5 t 2 + 10 t 2 ) (2t) 4 t 2 2 t 2 + 1 =0,

that is,

( 1 + t ) 2 φ(t)=0,
(3.5)

where

φ(t)= t 4 +10 t 2 2.

It is obvious that the equation

( 1 + t ) 2 =0
(3.6)

has no real root on the interval (0,1).

It is easy to find that the equation

φ(t)=0
(3.7)

has one positive real root. Moreover, it is not difficult to observe that φ(0)=2<0 and φ(1)=9>0. We can thus find that equation (3.7) has one distinct real root on the interval (0,1). So that equation (3.4) has only one real root t 0 given by t 0 =0.442890982868958 on the interval (0,1), and

g ( t ) max =g( t 0 )3.2817755127(3,4).
(3.8)

Now we prove g( t 0 ) is the root of equation (1.2). For this purpose, we consider the following nonlinear algebraic equation system:

{ φ ( t 0 ) = 0 , 2 t 0 2 + 1 u 0 2 = 0 , 4 t 0 2 v 0 2 = 0 , 4 ( 2 + t ) ( u 0 + v 0 ) 2 9 ( 2 t ) ( 1 + t ) 2 k = 0 .
(3.9)

It is easy to see that g( t 0 ) is also the solution of nonlinear algebraic equation system (3.9). If we eliminate the v 0 , u 0 and t 0 ordinal by the resultant (by using Lemma 2.9), then we get

29,648,323,021,629,456 ϕ 1 2 (k) ϕ 2 2 (k)=0,
(3.10)

where

ϕ 1 (k)=6,561 k 4 14,256 k 3 18,080 k 2 25,344k+20,736

and

ϕ 2 (k)=729 k 4 7,344 k 3 +8,800 k 2 13,056k+2,304.

The revised sign list of the discriminant sequence of ϕ 1 (k) is given by

[1,1,1,1].
(3.11)

The revised sign list of the discriminant sequence of ϕ 2 (k) is given by

[1,1,1,1].
(3.12)

So the number of sign changes in the revised sign list of (3.11) and (3.12) are both 2. Thus, by applying Lemma 2.10, we find that the equations

ϕ 1 (k)=0
(3.13)

and

ϕ 2 (k)=0
(3.14)

both have two distinct real roots. In addition, it is easy to find that

ϕ 1 ( 0 ) = 20 , 736 > 0 ; ϕ 2 ( 0 ) = 2 , 304 > 0 , ϕ 1 ( 1 ) = 30 , 383 < 0 ; ϕ 2 ( 1 ) = 8 , 567 < 0 , ϕ 1 ( 3 ) = 71 , 487 < 0 ; ϕ 2 ( 8 ) = 313 , 088 < 0

and

ϕ 1 (4)=397,312>0; ϕ 2 (9)=26,793>0.

We can thus find that equation (3.13) has two distinct real roots on the intervals

(0,1)and(3,4).

And equation (3.14) has two distinct real roots on the intervals

(0,1)and(8,9).

Hence, by (3.8), we can conclude that g( t 0 ) is the root of equation (1.2). The proof of Theorem 1.1 is thus completed. □

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Acknowledgements

The authors would like to thank the anonymous referees for their very careful reading and making some valuable comments which have essentially improved the presentation of this paper.

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Shi, SC., Wu, YD. An artificial proof of a geometric inequality in a triangle. J Inequal Appl 2013, 329 (2013). https://doi.org/10.1186/1029-242X-2013-329

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