Prove the inequality $\frac{a}{c+a-b}+\frac{b}{a+b-c}+\frac{c}{b+c-a}\ge{3}$ Let a, b, c be the three side lengths of a triangle. Prove that
$$\frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c}\geq 3$$
Under what conditions is equality obtained?

solution 1:
let $$b+c-a=x,a+c-b=y,a+b-c=z$$,
then
$$a=\dfrac{y+z}{2},b=\dfrac{x+z}{2},c=\dfrac{x+y}{2}$$
then
$$\sum\dfrac{b}{a+c-b}=\dfrac{1}{2}\left[\left(\dfrac{y}{x}+\dfrac{x}{y}\right)+\left(\dfrac{y}{z}+\dfrac{z}{y}\right)+\left(\dfrac{x}{z}+\dfrac{z}{x}\right)\right]\ge 3$$
solution 2:
by cauchy-Schwarz inequality we have
$$\sum\dfrac{a}{b+c-a}=\sum\dfrac{a^2}{a(b+c-a)}\ge\dfrac{(a+b+c)^2}{\sum a(b+c-a)}$$
$$\Longleftrightarrow (a+b+c)^2\ge\sum a(b+c-a)$$
$$\Longleftrightarrow 2a^2+2b^2+2c^2\ge ab+bc+ac$$
It's obivous.
solution 3:
since$a+b-c>0,b+c-a>0,a+c-b>0$
then we have
$$\sum\dfrac{2a}{b+c-a}=\sum\left(\dfrac{a+b-c}{b+c-a}+\dfrac{a-b+c}{b+c-a}\right)=\sum\left(\dfrac{a+b-c}{b+c-a}+\dfrac{b+c-a}{a+b-c}\right)\ge 6$$
