$a_1,a_2,a_3,b_1,b_2,b_3$ are positive real numbers, show: $\sqrt[3]{(a_1+b_1)(a_2+b_2)(a_3+b_3)} \geq \sqrt[3]{a_1a_2a_3} + \sqrt[3]{b_1b_2b_3}$

Question

The question says one only needs the AM-GM inequality, I've been stuck here for more than one hour.

$$(a_i + b_i) \gt a_i$$ and $$a_i + b_i \gt b_i$$

therefore,

$$ \sqrt[3]{(a_1+b_1)(a_2+b_2)(a_3+b_3)} > \sqrt[3]{a_1a_2a_3} $$

and

$$ \sqrt[3]{(a_1+b_1)(a_2+b_2)(a_3+b_3)} > \sqrt[3]{b_1b_2b_3} $$

so

$$ \sqrt[3]{(a_1+b_1)(a_2+b_2)(a_3+b_3)} > \frac{\sqrt[3]{a_1a_2a_3} + \sqrt[3]{b_1b_2b_3}}{2} $$

This was my best lower bound.

Answer 1

Divide both sides by $\sqrt[3]{(a_1+b_1)(a_2+b_2)(a_3+b_3)}$ and note \begin{align*} \sqrt[3]{\frac{a_1}{a_1+b_1}\frac{a_2}{a_2+b_2}\frac{a_3}{a_3+b_3}}&\leq\frac{1}{3}\left(\frac{a_1}{a_1+b_1}+\frac{a_2}{a_2+b_2}+\frac{a_3}{a_3+b_3}\right),\\ \sqrt[3]{\frac{b_1}{a_1+b_1}\frac{b_2}{a_2+b_2}\frac{b_3}{a_3+b_3}}&\leq\frac{1}{3}\left(\frac{b_1}{a_1+b_1}+\frac{b_2}{a_2+b_2}+\frac{b_3}{a_3+b_3}\right). \end{align*} Now add.

Answer 2

By Holder: $$(a_1+b_1)(a_2+b_2)(a_3+b_3) \geq \left(\sqrt[3]{a_1a_2a_3} + \sqrt[3]{b_1b_2b_3}\right)^3$$ and we are done!