Proof of inequality : $\frac{1}{a^3+b^3+abc}+\frac{1}{b^3+c^3+abc}+\frac{1}{c^3+a^3+abc}\le \frac{1}{abc}$

Question

Can we use AM-GM inequality proof method? I need a hint for proof.

Similar question: To prove $\sum_{cyc}\frac{1}{a^3+b^3+abc} \le \frac{1}{abc}$

If $a,b,c \in \mathbb R$ and $a,b,c > 0$, then $$ \frac{1}{a^3+b^3+abc}+\frac{1}{b^3+c^3+abc}+\frac{1}{c^3+a^3+abc}\le \frac{1}{abc} $$

Answer 1

By the rearrangement inequality, $a^3 + b^3 \geq a^2b + ab^2$. So we have $$\frac{1}{a^3+b^3+abc} \leq \frac{1}{a^2b+ab^2+abc} = \frac{1}{ab(a+b+c)}.$$ The LHS is hence at most $$\frac{1}{a+b+c} \left(\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ac}\right),$$ and the result follows since $$\frac{a+b+c}{abc} = \left(\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ac}\right).$$