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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.

Michael Rozenberg
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onlyme
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  • Try to cube the left side. If you defactorize this you will end up with $a_1 a_2 a_3 + b_1 b_2 b_3 + k$ where $k$ is the term with the rest of the multiplications and additions. $k$ is obviously greater or equal than zero since all other numbers are real and greater than zero. Thus your left side equals the right side cubed + a constant – Jan Jul 14 '15 at 14:52
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    If $A=(a_1,a_2,a_3),B=(b_1,b_2,b_3)$, your inequality follows from the super-additivity of the geometric mean: $$ GM(A+B)\geq GM(A)+GM(B). $$ – Jack D'Aurizio Jul 14 '15 at 16:26
  • I didn't know this property of the GM before – onlyme Jul 14 '15 at 16:32

2 Answers2

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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.

Kim Jong Un
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1

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!

Michael Rozenberg
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