Prove $\sum\limits_{cyc} \frac 1 {(a-b)^2} \ge \frac{9}{4}$ $a, b, c \in [0;2]$, Prove inequality: $$\frac1{(a-b)^2}+\frac1{(b-c)^2}+\frac1{(c-a)^2}\geq \frac94$$

I tried to:


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*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 $$\frac1{(a-b)^2}+\frac1{(b-c)^2}+\frac1{(c-a)^2}\geq \frac4{ab+bc+ca}$$


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 $$\frac{1}{b^2} + \frac{1}{(b-2)^2} + \frac{1}{2^2}, \ 0 < b < 2$$
