<s>[INST] Prove the inequality: $\frac{a}{c+a-b}+\frac{b}{a+b-c}+\frac{c}{b+c-a}\ge{3}$ Prove the inequality:
$\frac{a}{c+a-b}+\frac{b}{a+b-c}+\frac{c}{b+c-a}\ge{3}$
Where $a,b,c$ are sides of a triangle.
It is clear that $c+a-b$ is positive but how to use it?
 [/INST] $$a+b-c=x>0$$
$$c+a-b=y>0$$
$$b+c-a=z>0$$
The original inequality becomes
$$\frac{x+y}{2y}+\frac{x+z}{2x}+\frac{y+z}{2z}\geq 3$$
$$\frac{x}{2y}+\frac{z}{2x}+\frac{y}{2z}\geq \frac{3}{2}$$ which is trivial by AM-GM inequality
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