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(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 :)

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We prefer to avoid subjective titles. –  Pedro Tamaroff Sep 12 '12 at 19:26
    
Presumably you're looking at this in order to prepare for an upcoming competition. What use is it, then, if others solve the problem for you? –  Yuval Filmus Sep 12 '12 at 23:43
    
@YuvalFilmus I won't participate to a competition. It is only my pleasure to solve inequalities. If you want you can give me an idea. –  Iuli Sep 13 '12 at 6:46

1 Answer 1

up vote 2 down vote accepted

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.

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How do you know that : $$IA^2=\frac{(p-a) \cdot bc}{p} ? $$ Thanks :) –  Iuli Sep 13 '12 at 8:01
2  
Look it up: cut-the-knot.org/triangle/RelationsInTriangle.shtml –  ivan Sep 13 '12 at 8:18

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