SOLUTION. The inequality can be rewritten in the following form
$$\frac{4(a-b)^{2}}{a b}+\frac{4(c-a)(c-b)}{a c} \geq \frac{\left(a^{2}-b^{2}\right)^{2}}{\left(a^{2}+c^{2}\right)\left(c^{2}+b^{2}\right)}+\frac{\left(c^{2}-a^{2}\right)\left(c^{2}-b^{2}\right)}{\left(a^{2}+b^{2}\right)\left(a^{2}+c^{2}\right)}$$

WLOG, we may assume $c=\min (a, b, c)$. Then it's not too difficult to show that
$$\begin{array}{l} \frac{1}{a c} \geq \frac{(c+a)(c+b)}{\left(a^{2}+b^{2}\right)\left(c^{2}+b^{2}\right)} \\ \frac{4}{a b} \geq \frac{(a+b)^{2}}{\left(a^{2}+c^{2}\right)\left(b^{2}+c^{2}\right)} \end{array}$$
and the proof is completed. Equality holds for $a=b=c$.