Ta có \((b+c)^{2} \leq 2\left(b^{2}+c^{2}\right)\).

Suy ra \((b+c)^{2}+5 b c \leq 2\left(b^{2}+c^{2}\right)+5 b c=(b+2 c)(c+2 b)\)

Do đó \(\frac{a^{2}}{(b+c)^{2}+5 b c} \geq \frac{a^{2}}{(b+2 c)(c+2 b)}\)

Mặt khác \(a+b+c=a+\frac{b+2 c}{3}+\frac{c+2 b}{3} \geq 3 \sqrt[3]{a \cdot \frac{b+2 c}{3} \cdot \frac{c+2 b}{3}}\)

\(\Rightarrow \sqrt[3]{\frac{a^{2}}{(b+2 c)(c+2 b)}} \geq \sqrt[3]{3} \cdot \frac{a}{a+b+c} \Rightarrow \sqrt[3]{\frac{a^{2}}{(b+c)^{2}+5 b c}} \geq \sqrt[3]{3} \cdot \frac{a}{a+b+c}\)

Tương tự có \(\sqrt[3]{\frac{b^{2}}{(c+a)^{2}+5 c a}} \geq \sqrt[3]{3} \cdot \frac{b}{a+b+c} ; \sqrt[3]{\frac{c^{2}}{(a+b)^{2}+5 a b}} \geq \sqrt[3]{3} \cdot \frac{c}{a+b+c}\)

Do vậy \(P \geq \sqrt[3]{3}\). Dấu \(=\) xảy ra khi \(a=b=c\).

Vậy \(\min P=\sqrt[3]{3}\)