Inequality. $3\left(IA^2+IB^2+IC^2\right) \geq AB^2+BC^2+CA^2$ (Korea 1998) Let $I$ be the incenter of a triangle $ABC$.Prove that: 

$$3\left(IA^2+IB^2+IC^2\right) \geq AB^2+BC^2+CA^2.$$ 

Please help me to improve this kind of inequalities. 
Thanks :) 

First note that $IA^2=(p-a)bc/p$ and similarly for $IB, IC$. Substitute to get$$
3abc(\frac{p-a}{ap}+\frac{p-b}{bp}+\frac{p-c}{cp})-a^2-b^2-c^2>=0
$$
Replace $p$ with $(a+b+c)/2$ and get rid of the fractions to arrive at:$$
-a^3 + 2 a^2 b + 2 a b^2 - b^3 + 2 a^2 c - 9 a b c + 2 b^2 c + 
 2 a c^2 + 2 b c^2 - c^3>=0
$$
Substitute $a=x+y, b=x+z, c=y+z$ because they are triangle sides and simplify to get:$$
x^3 - x^2 y - x y^2 + y^3 - x^2 z + 3 x y z - y^2 z - x z^2 - 
 y z^2 + z^3>=0
$$
This I think you know how to prove.
