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Algebra Difficulty 6.4 National olympiad Prove it

22. Prove the following inequality
(2)n(a1+a2)(a2+a3)⋯(an+a1)≤(a1+a2+a3)(a2+a3+a4)⋯(an+a1+a2)(\sqrt{2})^{n}\left(a_{1}+a_{2}\right)\left(a_{2}+a_{3}\right) \cdots\left(a_{n}+a_{1}\right) \leq\left(a_{1}+a_{2}+a_{3}\right)\left(a_{2}+a_{3}+a_{4}\right) \cdots\left(a_{n}+a_{1}+a_{2}\right)

where a1,a2,⋯ ,ana_{1}, a_{2}, \cdots, a_{n} are any positive real numbers (Russia MO)

Solution

Prove: According to the following inequalities
(a1+a2+a3)2≥(2a1+a2)(a2+2a3)(2a1+a2)(2a2+a1)=2a12+2a22+5a1a2≥2(a1+a2)2\begin{array}{l} \left(a_{1}+a_{2}+a_{3}\right)^{2} \geq\left(2 a_{1}+a_{2}\right)\left(a_{2}+2 a_{3}\right) \\ \left(2 a_{1}+a_{2}\right)\left(2 a_{2}+a_{1}\right)=2 a_{1}^{2}+2 a_{2}^{2}+5 a_{1} a_{2} \geq 2\left(a_{1}+a_{2}\right)^{2} \end{array}

we can immediately derive the result, because
2n∏cyc(a1+a2)2≤∏cyc(2a1+a2)(2a2+a1)=∏cyc(2a1+a2)(a2+2a3)≤∏cyc(a1+a2+a3)2\begin{array}{l} 2^{n} \prod_{c y c}\left(a_{1}+a_{2}\right)^{2} \leq \prod_{c y c}\left(2 a_{1}+a_{2}\right)\left(2 a_{2}+a_{1}\right)=\prod_{c y c}\left(2 a_{1}+a_{2}\right)\left(a_{2}+2 a_{3}\right) \\ \leq \prod_{c y c}\left(a_{1}+a_{2}+a_{3}\right)^{2} \end{array}

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.