# Prove that $\frac{1}{a^3+b^3+abc}+\frac{1}{b^3+c^3+abc}+\frac{1}{c^3+a^3+abc} \leq \frac{1}{abc}.$ [duplicate]

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.

## marked as duplicate by Batominovski, Davide Giraudo, Alex Provost, Rory Daulton, wythagorasDec 30 '15 at 20:56

• @zz20s The Arithmetic-Geometric Mean Inequality. For all $a_1,a_2,\ldots,a_k\ge 0$ we have $a_1+a_2+\cdots+a_k\ge k\sqrt{a_1a_2\cdots a_k}$ with equality if and only if $a_1=a_2=\cdots =a_k$. – user236182 Dec 30 '15 at 15:23
By AM-GM or Muirhead, $\frac{abc}{a^3+b^3+abc}\leq \frac{abc}{a^2b+ab^2+abc}=\frac{c}{a+b+c}$.
• @Btominovski How did you use AM-GM? Didn't you just use the fact that $a^3+b^3 \geq a^2b+ab^2$? – user19405892 Dec 30 '15 at 15:04
• @user19405892 It should be $a^3+b^3\ge a^2b+ab^2$. Note that by AM-GM $a^3+b^3+b^3\ge 3ab^2$ and $a^3+a^3+b^3\ge 3a^2b$ and add these two inequalities. – user236182 Dec 30 '15 at 15:05