Prove that $\frac{1}{a^3+b^3+abc}+\frac{1}{b^3+c^3+abc}+\frac{1}{c^3+a^3+abc} \leq \frac{1}{abc}.$ 
Let $a,b,$ and $c$ be positive real numbers. Prove that $$\dfrac{1}{a^3+b^3+abc}+\dfrac{1}{b^3+c^3+abc}+\dfrac{1}{c^3+a^3+abc} \leq \dfrac{1}{abc}.$$

This question seems hard since we aren't given any information on $a,b,c$ except that they are positive. I tried cross multiplying by $abc$ and proving that $$\dfrac{abc}{a^3+b^3+abc}+\dfrac{abc}{b^3+c^3+abc}+\dfrac{abc}{c^3+a^3+abc} \leq 1.$$ Then we use AM-GM on the denominator. This didn't  seem to help me.

By AM-GM or Muirhead, $\frac{abc}{a^3+b^3+abc}\leq \frac{abc}{a^2b+ab^2+abc}=\frac{c}{a+b+c}$.
