Prove the inequality: $\frac{a}{c+a-b}+\frac{b}{a+b-c}+\frac{c}{b+c-a}\ge{3}$ [duplicate]
Prove the inequality:
$\frac{a}{c+a-b}+\frac{b}{a+b-c}+\frac{c}{b+c-a}\ge{3}$
Where $a,b,c$ are sides of a triangle.
It is clear that $c+a-b$ is positive but how to use it?
Solution 1:
$$a+b-c=x>0$$ $$c+a-b=y>0$$ $$b+c-a=z>0$$ The original inequality becomes $$\frac{x+y}{2y}+\frac{x+z}{2x}+\frac{y+z}{2z}\geq 3$$ $$\frac{x}{2y}+\frac{z}{2x}+\frac{y}{2z}\geq \frac{3}{2}$$ which is trivial by AM-GM inequality
Solution 2:
Hint:
use this substitution $$a=y+z,\quad b=x+z,\quad c=x+y$$