Prove the inequality $\frac{a}{c+a-b}+\frac{b}{a+b-c}+\frac{c}{b+c-a}\ge{3}$

Question

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?

Answer 1

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 $$\sum\dfrac{2a}{b+c-a}=\sum\left(\dfrac{a+b-c}{b+c-a}+\dfrac{a-b+c}{b+c-a}\right)=\sum\left(\dfrac{a+b-c}{b+c-a}+\dfrac{b+c-a}{a+b-c}\right)\ge 6$$