<s>[INST] Let $a$, $b$ and $c$ be the three sides of a triangle. Show that $\frac{a}{b+c-a}+\frac{b}{c+a-b} + \frac{c}{a+b-c}\geqslant3$. 
Let $a$, $b$ and $c$ be the three sides of a triangle.
Show that $$\frac{a}{b+c-a}+\frac{b}{c+a-b} + \frac{c}{a+b-c}\geqslant3\,.$$

A full expanding results in:
$$\sum_{cyc}a(a+b-c)(a+c-b)\geq3\prod_{cyc}(a+b-c),$$ or
$$\sum_{cyc}(a^3-ab^2-ac^2+2abc)\geq\sum_{cyc}(-3a^3+3a^2b+3a^2c-2abc),$$ but it becomes very ugly.
 [/INST] Suppose that $S>0$. Then for $x\in(0,S/2)$, the function $f(x)=\frac{x}{S-2x}$ is convex. Thus, by Jensen's inequality and with $S=a+b+c$, we have
$$
\frac{1}{3}f(a)+\frac{1}{3}f(b)+\frac{1}{3}f(c)\geq f\left(\frac{1}{3}(a+b+c)\right)=\frac{S/3}{S-2S/3}=1.
$$
This is equivalent to $f(a)+f(b)+f(c)\geq 3$, which is your inequality.
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