\(\frac{a^4}{b+c}+\frac{b^4}{c+a}+\frac{c^4}{a+b}=\frac{a^6}{a^2b+a^2c}+\frac{b^6}{b^2a+b^2c}+\frac{c^6}{c^2a+c^2b}\ge\frac{\left(a^3+b^3+c^3\right)^2}{ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)}\ge\frac{\left(a^3+b^3+c^3\right)^2}{2\left(a^3+b^3+c^3\right)}=\frac{a^3+b^3+c^3}{2}\)
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