Let $a\,b$ and $c$ be the sides of a triangle. Prove that $\displaystyle \frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c}\ge 3$.

Let $a\,b$ and $c$ be the sides of a triangle.

Prove that $\displaystyle \frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c}\ge 3$.

My solution:

$\displaystyle \frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c}$

$\displaystyle =\frac{a^2}{a(b+c)-a^2}+\frac{b^2}{b(a+c)-b^2}+\frac{c^2}{c(a+b)-c^2}$

$\displaystyle \ge 4\left(\left(\frac{a}{b+c}\right)^2+\left(\frac{b}{c+a}\right)^2+\left(\frac{c}{a+b}\right)^2\right)\,\,\text{since}\,\,a^2+\frac{(b+c)^2}{4}\ge a(b+c)$

$\displaystyle \ge 4\left(\frac{\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)^2}{3}\right)\,\,\text{since}\,\,3(x^2+y^2+z^2)\ge (x+y+z)^2$

$\displaystyle \ge 4\left(\frac{\left(\frac{3}{2}\right)^2}{3}\right)\,\,\text{from the Nesbitt's inequality}$\displaystyle \ge 3$