# 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 $\frac{a}{b+c-a}+\frac{b}{c+a-b} + \frac{c}{a+b-c}\geqslant3$

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$a, b, c$ are sides of a triangle iff there exists positive reals $x, y, z$ s.t. $a=x+y, b=y+z, c = z+x$. In terms of these variables, the inequality is
$$\sum_{cyc} \frac{a}{b+c-a} = \sum_{cyc} \frac{x+y}{2z} \ge 3$$

Now the last is easy to show with AM-GM of all $6$ terms.
$$\sum_{cyc} \frac{x+y}{2z} = \frac12\left(\frac{x}z+\frac{y}z+\frac{y}x+\frac{z}x+\frac{z}y+\frac{x}y \right) \ge \frac12\left(6\sqrt[6]{\frac{x}z\cdot\frac{y}z\cdot\frac{y}x\cdot\frac{z}x\cdot\frac{z}y\cdot\frac{x}y} \right) = 3$$

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.