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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.

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