\(\frac{a^3}{b+c}+\frac{b^3}{a+c}+\frac{c^3}{a+b}=\frac{a^4}{ab+ac}+\frac{b^4}{ab+bc}+\frac{c^4}{ac+bc}\ge\frac{\left(a^2+b^2+c^2\right)^2}{2\left(ab+bc+ca\right)}\ge\frac{\left(ab+bc+ca\right)^2}{2\left(ab+bc+ca\right)}=\frac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c=1\)