The sum of the numbers $a,b$ and $c$ is zero, and their product is negative. Prove that the number $\frac{a^2+b^2}{c}+\frac{b^2+c^2}{a}+\frac{c^2+a^2}{b}$ is positive.
Problem
Source: Caucasus 2015 8.4
Tags: algebra, inequalities, three variable inequality
26.04.2019 16:21
Let $a,b,c$ be reals such that $a+b+c=0$ and $abc<0.$ Prove that $$\frac{a^2+b^2}{c}+\frac{b^2+c^2}{a}+\frac{c^2+a^2}{b}>0.$$Proof of Zhangyanzong:
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27.04.2019 04:57
parmenides51 wrote: The sum of the numbers $a,b$ and $c$ is zero, and their product is negative. Prove that the number $\frac{a^2+b^2}{c}+\frac{b^2+c^2}{a}+\frac{c^2+a^2}{b}$ is positive. Solution. Since $a+b+c=0$ and $abc<0$, we deduce that \begin{align*}&\frac{a^2+b^2}{c}+\frac{b^2+c^2}{a}+\frac{c^2+a^2}{b}\\ =&\frac{(a+b)^2-2ab}{c}+\frac{(b+c)^2-2bc}{a}+\frac{(c+a)^2-2ca}{b}\\ =&\frac{c^2-2ab}{c}+\frac{a^2-2bc}{a}+\frac{b^2-2ca}{b}\\ =&c-\frac{2abc}{c^2}+a-\frac{2abc}{a^2}+b-\frac{2abc}{b^2}\\ =&-2abc\left(\frac{1}{c^2}+\frac{1}{a^2}+\frac{1}{b^2}\right)>0. \end{align*}The result follows. $\blacksquare$
27.04.2019 13:23
28.07.2019 03:56
WLOG $a, b>0$ $\frac{a^2+b^2}{c}+\frac{b^2+c^2}{a}+\frac{c^2+a^2}{b}=\frac{b^2+c^2}{a}+\frac{c^2+a^2}{b}-\frac{a^2+b^2}{a+b}>\frac{(b^2+c^2)+(c^2+a^2)-(a^2+b^2)}{a+b}=\frac{2 c^2}{a+b}>0$