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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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    @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.2012-09-13

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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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    Look it up: http://www.cut-the-knot.org/triangle/RelationsInTriangle.shtml2012-09-13