Giả sử $c=min(a,b,c)$ vì $a,b,c\geq 0 \Rightarrow ab+bc+ca \geq ab$
$\frac{1}{(b-c)^2}\geq \frac{1}{b^2}$ $ \frac{1}{(a-c)^2}\geq \frac{1}{a^2}$
Vậy ta chỉ cần chứng minh $ab\left ( \frac{1}{(a-b)^2}+\frac{1}{b^2}+\frac{1}{a^2} \right )\geq 4$
$\Leftrightarrow \frac{ab}{(a-b)^2}+\frac{a}{b}+\frac{b}{a}-4\geq 0\Leftrightarrow \frac{ab}{(a-b)^2}+\frac{(a-b)^2}{ab}-2\geq 0 \Leftrightarrow \left ( \sqrt{\frac{ab}{(a-b)^2}} -\sqrt{\frac{(a-b)^2}{ab}}\right )^2\geq 0$