SOLUTION. WLOG, assume that $a \geq b \geq c$. Denote
$$f(a, b, c)=\frac{a^{2}}{b^{2}+c^{2}}+\frac{b^{2}}{c^{2}+a^{2}}+\frac{c^{2}}{a^{2}+b^{2}}+\frac{27(a+b+c)^{2}}{a^{2}+b^{2}+c^{2}} .$$

We will prove that $f(a, b, c) \geq f\left(a, \sqrt{b^{2}+c^{2}}, 0\right)$. Indeed
$$\begin{array}{c} f(a, b, c)-f\left(a, \sqrt{b^{2}+c^{2}}, 0\right) \\ =\frac{b^{2}}{c^{2}+a^{2}}+\frac{c^{2}}{a^{2}+b^{2}}-\frac{b^{2}+c^{2}}{a^{2}}+\frac{27(a+b+c)^{2}-27\left(a+\sqrt{b^{2}+c^{2}}\right)^{2}}{a^{2}+b^{2}+c^{2}} \\ \geq-b^{2} c^{2}\left(\frac{1}{a^{2}\left(a^{2}+b^{2}\right)}+\frac{1}{a^{2}\left(a^{2}+c^{2}\right)}\right)+\frac{54 a\left(b+c-\sqrt{b^{2}+c^{2}}\right)}{a^{2}+b^{2}+c^{2}} \\ \geq-b^{2} c^{2}\left(\frac{1}{a^{2}\left(a^{2}+b^{2}\right)}+\frac{1}{a^{2}\left(a^{2}+c^{2}\right)}\right)+\frac{54 b^{2} c^{2}}{4 b c\left(a^{2}+b^{2}+c^{2}\right)} \end{array}$$

Moreover, because
$$\begin{array}{c} \frac{3}{a^{2}} \geq \frac{54}{4 b c} \\ \frac{1}{a^{2}\left(a^{2}+b^{2}\right)}+\frac{1}{a^{2}\left(a^{2}+c^{2}\right)} \leq \frac{3}{a^{2}+b^{2}+c^{2}} \end{array}$$

We infer that $f(a, b, c) \geq f\left(a, \sqrt{b^{2}+c^{2}}, 0\right)$. Denote $t=\sqrt{b^{2}+c^{2}}$, then we have
$$\begin{array}{l} f(a, t, 0)=\frac{a^{2}}{t^{2}}+\frac{t^{2}}{a^{2}}+\frac{27(a+t)^{2}}{a^{2}+t^{2}}=-2+\frac{\left(a^{2}+t^{2}\right)^{2}}{a^{2} t^{2}}+\frac{54 a t}{a^{2}+t^{2}}+27 \\ \quad=\frac{\left(a^{2}+t^{2}\right)^{2}}{a^{2} t^{2}}+\frac{27 a t}{a^{2}+t^{2}}+\frac{27 a t}{a^{2}+t^{2}}+25 \geq 3 \sqrt[3]{27 \cdot 27}+25=52 \end{array}$$

This ends the proof. The equality holds for $(a, b, c) \sim\left(\frac{-3 \pm \sqrt{5}}{2}, 1,0\right)$.