$$\text { 77. } \begin{array}{l} \frac{a^{3}}{b+2 c}+\frac{b^{3}}{c+2 a}+\frac{c^{3}}{a+2 b}=\frac{a^{4}}{a b+2 a c}+\frac{b^{4}}{b c+2 a b}+\frac{c^{4}}{a c+2 b c} \geqslant \\ \frac{\left(a^{2}+b^{2}+c^{2}\right)^{2}}{a b+2 a c+b c+2 a b+a c+2 b c}= \\ \left(a^{2}+b^{2}+c^{2}\right) \cdot \frac{a^{2}+b^{2}+c^{2}}{3(a b+b c+a c)} \end{array}$$

Given that $a^{2}+b^{2}+c^{2} \geqslant a b+b c+a c$, therefore
$$\frac{a^{3}}{b+2 c}+\frac{b^{3}}{c+2 a}+\frac{c^{3}}{a+2 b} \geqslant \frac{a^{2}+b^{2}+c^{2}}{3}$$