Ta có: \(1=a^2+b^2+c^2\ge ab+bc+ca\).
\(P=\dfrac{a^3}{b+2c}+\dfrac{b^3}{c+2a}+\dfrac{c^3}{a+2b}=\dfrac{a^4}{ab+2ca}+\dfrac{b^4}{bc+2ab}+\dfrac{c^4}{ca+2bc}\)
\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{3\left(ab+bc+ca\right)}=\dfrac{1}{3\left(ab+bc+ca\right)}\ge\dfrac{1}{3}\)
Dấu \(=\) xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\).