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let $$0\le a_{1}\le a_{2}\le \cdots\le a_{3n}$$ show that $$\left(a_{1}+a_{2}+a_{3}+\cdots+a_{3n-1}+a_{3n}\right)^3\ge 27n^2\sum_{k=1}^{n}a_{k}a_{n+k}a_{2n+k}$$

we know when $n=1$，this is $$(a+b+c)^3\ge 27abc$$ this is AM-GM inequality. because $$a+b+c\ge 3\sqrt[3]{abc}$$ $$\Longrightarrow (a+b+c)^3\ge 27abc$$

when $n=2$, then we have $$(a_{1}+a_{2}+a_{3}+\cdots+a_{6})^3\ge 108(a_{1}a_{3}a_{5}+a_{2}a_{4}a_{6})$$ this inequality maybe can use AM-GM inequality to solve it.and For general How prove it? Thank you

math110
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1 Answers1

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HINT

Group into 3 terms. So

$$ (a_1+a_2+\cdots+a_{3n})^3 >= (a_1+a_{n+1}+a_{2n+1})^3+\cdots+(a_n+a_{2n}+a_{3n})^3 $$

Then use AM-GM for RHS terms. Are you sure the RHS is $27n^2$ and not $27n$ ?

dear chap
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