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Let $(a_1,\space a_2,\space \cdots, \space a_n) \in \mathbb R^n_+$ such that $\displaystyle \prod^n_{i=1 }a_i = 1$. Prove that $$\displaystyle \prod^n_{i=1} (1+a_i^2) \le \cfrac {2^n}{n^{2n-2}}\left (\sum^n_{i=1} a_i\right)^{2n-2}$$

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    $a_1,\cdots ,a_n$ are positive real numbers,I guess?2012-10-29
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    @yzhao no, they are just real numbers2012-10-29
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    @FOla Yinka:If $4\vert n$,consider a sequence $(a_1,\cdots,a_n)=(1,1,\cdots,-1,-1)$ with $n/2$ 1s and $n/2$ -1s.Then $LHS>0$ and $RHS=0$.In other words,the inequality fails!2012-10-29
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    Assuming the numbers are positive, if their product is $1$ their sum must be at least $n$.2012-10-29
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    *there is no other way I could have written this question*... Hmmm, at least you could have included what you tried and where your tries failed, as they say one should do on this site...2012-10-29
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    @did I've read the FAQ, I don't see that as a prerequisite for asking questions. This is not a homework.2012-10-29
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    I apologize to everyone, $a_1,\cdots ,a_n$ are positive.2012-10-29
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    While including what one tried and where these tries failed might not be mandatory for not-homework questions, it is definitely recommended. Not doing so (and, in the present case, *refusing* to do so) is a clear message to (at least some) potential answerers. Are you sure you want to convey this message?2012-10-29
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    I tried substituting $a_i=b_i^n$ and homogenizing to get $\prod (\prod b_i^2 + b_i^{2n})/2) \leq (\sum (b_i^n)/n)^{2n-2} \prod b_i^2$, which is then an inequality of symmetric polynomials. It does not seem to follow readily from Muirhead's, and I'm not so versed in more advanced ones.2012-11-11
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    This question is hard enough that including failed partial attempts would probably not be helpful.2012-11-16
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    Is there a good reason to believe that it is true?2012-11-16

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