Kvant Math Problem 7

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Problem

$a$, $b$, $c$ are the sides of a triangle. Prove that

$$\frac{a}{b+c-a} + \frac{b}{c+a-b} + \frac{c}{a+b-c} \geq 3.$$

S. T. Berkolayko

Exploration

Let

$$S=\frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c}.$$

Since $a,b,c$ are sides of a triangle, each denominator is positive by the triangle inequalities.

A natural test is the equilateral triangle. For $a=b=c$,

$$S=1+1+1=3.$$

This suggests that $3$ may be the minimum.

To understand the structure, introduce the standard substitution

$$a=y+z,\qquad b=z+x,\qquad c=x+y,$$

where $x,y,z>0$. Then

$$b+c-a=(z+x)+(x+y)-(y+z)=2x,$$

and cyclically

$$c+a-b=2y,\qquad a+b-c=2z.$$

Hence

$$S=\frac{y+z}{2x}+\frac{z+x}{2y}+\frac{x+y}{2z}.$$

Expanding,

$$2S=\left(\frac yx+\frac xy\right) +\left(\frac zy+\frac yz\right) +\left(\frac xz+\frac zx\right).$$

Now each pair satisfies

$$\frac yx+\frac xy\ge 2,$$

because

$$\frac yx+\frac xy-2 =\frac{(x-y)^2}{xy}\ge0.$$

Summing the three inequalities gives $2S\ge6$, hence $S\ge3$.

The only point that could hide an error is the substitution $a=y+z$, $b=z+x$, $c=x+y$. One must justify that every triangle admits such positive $x,y,z$.

Problem Understanding

We must prove that for any triangle with side lengths $a,b,c$,

$$\frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c}\ge3.$$

This is a Type B problem, a pure proof.

The core difficulty is to exploit the triangle inequalities hidden in the denominators. The expression becomes much more symmetric after replacing the side lengths by

$$a=y+z,\qquad b=z+x,\qquad c=x+y.$$

After this transformation, the inequality reduces to three applications of the elementary inequality

$$t+\frac1t\ge2.$$

Proof Architecture

Every triangle can be represented in the form

$$a=y+z,\qquad b=z+x,\qquad c=x+y$$

with $x,y,z>0$; this follows from the triangle inequalities by setting

$$x=\frac{b+c-a}{2},\quad y=\frac{c+a-b}{2},\quad z=\frac{a+b-c}{2}.$$

Under this substitution, the denominators simplify to

$$b+c-a=2x,\quad c+a-b=2y,\quad a+b-c=2z.$$

After substitution,

$$2S= \left(\frac yx+\frac xy\right) +\left(\frac zy+\frac yz\right) +\left(\frac xz+\frac zx\right).$$

For any positive numbers $u,v$,

$$\frac uv+\frac vu\ge2,$$

because the difference from $2$ equals

$$\frac{(u-v)^2}{uv}.$$

Adding the three resulting inequalities yields $2S\ge6$, hence $S\ge3$.

The lemma most likely to fail under scrutiny is the first one, namely the justification that the substitution produces positive variables for every triangle.

Solution

Define

$$x=\frac{b+c-a}{2},\qquad y=\frac{c+a-b}{2},\qquad z=\frac{a+b-c}{2}.$$

Since $a,b,c$ are the sides of a triangle,

$$b+c-a>0,\qquad c+a-b>0,\qquad a+b-c>0,$$

so $x,y,z$ are positive.

From the definitions,

$$y+z =\frac{c+a-b+a+b-c}{2} =a,$$

$$z+x =\frac{a+b-c+b+c-a}{2} =b,$$

and

$$x+y =\frac{b+c-a+c+a-b}{2} =c.$$

Hence

$$a=y+z,\qquad b=z+x,\qquad c=x+y.$$

Let

$$S=\frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c}.$$

Using the relations above,

$$b+c-a=(z+x)+(x+y)-(y+z)=2x,$$

and similarly

$$c+a-b=2y,\qquad a+b-c=2z.$$

Therefore

$$S=\frac{y+z}{2x}+\frac{z+x}{2y}+\frac{x+y}{2z},$$

$$2S= \left(\frac yx+\frac zx\right) +\left(\frac zy+\frac xy\right) +\left(\frac xz+\frac yz\right).$$

Rearranging terms,

$$2S$$