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.

Solution 1:

$a, b, c$ 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. $$\sum_{cyc} \frac{x+y}{2z} = \frac12\left(\frac{x}z+\frac{y}z+\frac{y}x+\frac{z}x+\frac{z}y+\frac{x}y \right) \ge \frac12\left(6\sqrt[6]{\frac{x}z\cdot\frac{y}z\cdot\frac{y}x\cdot\frac{z}x\cdot\frac{z}y\cdot\frac{x}y} \right) = 3$$

Solution 2:

Set $b+c-a=x, a+c-b=y$ and $a+b-c=z$
Due to triangular inequality, $x,y,z$ are positive. Hence we can apply the AM-GM inequality. We have $x+y=b+c-a+a+c-b=2c \implies c = \frac{x+y}{2}$
Similarly $a = \frac{y+z}{2}$ and $b = \frac{z+x}{2}$
Now we have, $$\frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c}=\frac{y+z}{2x}+\frac{z+x}{2y}+\frac{x+y}{2z}$$ $$\Leftrightarrow 2\left ( \frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c} \right )=\left ( \frac{x}{y}+\frac{y}{x} \right )+\left ( \frac{y}{z}+\frac{z}{y} \right )+\left ( \frac{z}{x}+\frac{x}{z} \right )\geq 6$$ Hence $\ $ $\frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c}\geq 3$