Let $f(a, b, c) = \frac{a}{a^2 + b^3 + c^3} + \frac{b}{a^3 + b^2 + c^3} + \frac{c}{a^3 + b^3 + c^2}$ and $g(a, b, c) = a(a^2 + b^3 + c^3) + b(a^3 + b^2 + c^3) + c(a^3 + b^3 + c^2)$. Observe that the Cauchy-Schwarz Inequality gives $f(a, b, c) \cdot g(a, b, c) \ge (a+b+c)^2$.

Observe that $g(a, b, c) = (a+b+c)(a^3+b^3+c^3)$ since $a^3+b^3+c^3 = a^4+b^4+c^4$ and therefore it suffices to show that $a+b+c \ge a^3+b^3+c^3$. Again by the Cauchy-Schwarz Inequality we have $(a+b+c)(a^3+b^3+c^3) \ge (a^2+b^2+c^2)^2$ and $(a^2+b^2+c^2)(a^4+b^4+c^4) \ge (a^3+b^3+c^3)^2$. As $a^3+b^3+c^3 = a^4+b^4+c^4$, the result follows from the last two inequalities.