# Prove $\sum \frac 1 {(a-b)^2} \ge \frac{9}{4}$

$a, b, c \in [0;2]$, Prove inequality: $$\frac1{(a-b)^2}+\frac1{(b-c)^2}+\frac1{(c-a)^2}\geq \frac94$$

I tried to:

• Use AM-GM: $$LHS \ge \frac{(1+1+1)^2}{(a-b)^2+(b-c)^2+(c-a)^2}=\frac9{2(a^2+b^2+c^2 -ab-bc-ca)}$$ I am proving $a^2 + b^2+c^2 - ab - bc - ca \le 2$ but stuck
• Transform, we have: $$(a-b)+(b-c)+(c-a)=0 \implies LHS=\begin{vmatrix}\frac1{a-b}+\frac1{b-c}+\frac1{c-a}\end{vmatrix}$$ stuck here too. One more inequality (may be useful) is $$\frac1{(a-b)^2}+\frac1{(b-c)^2}+\frac1{(c-a)^2}\geq \frac4{ab+bc+ca}$$
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WLOG we can assume $a < b < c$. Since multiplying $a,b,c$ all by the same constant $> 1$ decreases the left side, we may assume $c = 2$. Since decreasing $a$ and $b$ by the same positive number decreases the left side, we may assume $a=0$. Now there's only one variable left, and I'll leave it to you to minimize $$\frac{1}{b^2} + \frac{1}{(b-2)^2} + \frac{1}{2^2}, \ 0 < b < 2$$

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So simple !!!!!! – Xeing Feb 14 '13 at 8:17
+1. Marvelous and pretty short. – Felix Marin Jul 31 '14 at 19:24