SolutION. Applying Cauchy-Schwarz, we obtain
$$\left(\sum_{c y c} \frac{a}{b^{2}+c^{2}}\right)\left(\sum_{c y c} a\left(b^{2}+c^{2}\right)\right) \geq(a+b+c)^{2}$$

It remains to prove that
$$\frac{(a+b+c)^{2}}{a b(a+b)+b c(b+c)+c a(c+a)} \geq \frac{4}{5} \cdot \frac{a^{2}+b^{2}+c^{2}+3(a b+b c+c a)}{a b(a+b)+b c(b+c)+c a(c+a)+2 a b c}$$

Let $S=\sum_{\text {cyc }} a^{2}, P=\sum_{\text {cyc }} a b$ and $Q=\sum_{\text {cyc }} a b(a+b)$. The above inequality becomes
$$\frac{5(S+2 P)}{Q} \geq \frac{4(S+3 P)}{Q+2 a b c} \Leftrightarrow S Q+10 a b c S+20 a b c P \geq 2 P Q$$

Clearly, we have
$$\begin{array}{l} P Q=\sum_{s y m} a^{2} b^{2}(a+b)+2 a b c(S+P) \\ S Q \geq \sum_{s y m} a b(a+b)\left(a^{2}+b^{2}\right) \geq 2 \sum_{\text {sym }} a^{2} b^{2}(a+b) \end{array}$$

This finishes the proof. Equality holds for $a=b, c=0$ up to permutations.