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Let $a,b,c$ be the lengths of the sides of a triangle. Prove that, $$a^2b(a-b)+b^2c(b-c)+c^2a(c-a) \geq 0.$$

Attempt

Using a ravi substitution I said let $a = x+y, b = y+z, c = z+x$. Then we have $$(x+y)^2(y+z)(x-z)+(y+z)^2(x+z)(y-x)+(x+z)^2(x+y)(z-y) = 2 x y^3+2 x^3 z + 2 y z^3-2 x^2 y z-2 x y^2 z-2 x y z^2.$$ Then I thought of using the rearrangement in equality, but I couldn't see how to make that work.

Jacob Willis
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1 Answers1

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This and some other versions are very common in IMO and local math competitions. The general technique is to make them square form so that it becomes $Ax^2+By^2+Cz^2$. For this one the following is equivalent to your question:

$$ \frac12((a-b)^2(a+b-c)(b+c-a)+(b-c)^2(b+c-a)(a+c-b)+(c-a)^2(a+c-b)(a+b-c))\geq 0. $$

user64066
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  • you skipped one point, namely that you have to show $A, B, C$ are all nonnegative. (which is implied by the triangle inequality) – Jason S Jan 03 '16 at 21:07