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Inequality 54 (George Basdekis)

Problem:

Let $a,b$ and $c$ be real numbers. Prove the Inequality

$\displaystyle\frac{1}{(a-b)^2}+\frac{1}{(b-c)^2}+\frac{1}{(c-a)^2}\geq\frac{9}{2(a^2+b^2+c^2)}.$

Solution:

Without loss of generality, let us assume that $a>b>c$. Then according to the AM-GM Inequality we have that

\displaystyle\begin{aligned}\frac{1}{(a-b)^2}+\frac{1}{(b-c)^2}\geq\frac{2}{(a-b)(b-c)}&\geq\frac{2\cdot 4}{[(a-b)+(b-c)]^2}\\&=\frac{8}{(a-c)^2}.\end{aligned}

Therefore, it holds that

$\displaystyle\frac{1}{(a-b)^2}+\frac{1}{(b-c)^2}+\frac{1}{(c-a)^2}\geq\frac{9}{(c-a)^2}.$

It remains now to prove that

$\displaystyle \frac{9}{(c-a)^2}\geq\frac{9}{2(a^2+b^2+c^2)},$

which reduces to the obvious, after expansions, Inequality $(a+c)^2+2b^2\geq 0.$ Equality holds if and only if $(a,b,c)\sim (-b,b,0)$ or any cyclic permutations of the previous equality Q.E.D.