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From brilliant.org, the Chebyshev's inequality, a generalization of the rearrangement inequality, states that for every nonincreasing sequences $\{a_i\}_{i=1}^n$ and $\{b_i\}_{i=1}^n$,

$$ \frac{1}{n}\sum_{i=1}^{n}a_ib_i \geq \left(\frac{1}{n}\sum_{i=1}^{n}a_i\right)\left(\frac{1}{n}\sum_{i=1}^{n}b_i\right) \geq \frac{1}{n}\sum_{i=1}^{n}a_ib_{n-i+1}\,. $$ As a special case when $a_i = b_i = x_i$, we have $$ \frac{1}{n}\sum_{i=1}^{n}x^2_i \geq \left(\frac{1}{n}\sum_{i=1}^{n}x_i\right)^2 \geq \frac{1}{n}\sum_{i=1}^{n}x_ix_{n-i+1}\,. $$ Clearly, knowing that $a\leq \frac{1}{2}(a+b)\leq b$, $$ \frac{1}{n}\sum_{i=1}^{n}x^2_i \geq \frac{1}{2}\left(\frac{1}{n}\sum_{i=1}^{n}x^2_i + \frac{1}{n}\sum_{i=1}^{n}x_ix_{n-i+1}\right) \geq \frac{1}{n}\sum_{i=1}^{n}x_ix_{n-i+1}\,. $$ Question: When does the following inequality hold? $$ \frac{1}{2}\left(\frac{1}{n}\sum_{i=1}^{n}x^2_i + \frac{1}{n}\sum_{i=1}^{n}x_ix_{n-i+1}\right) \leq \left(\frac{1}{n}\sum_{i=1}^{n}x_i\right)^2 $$

UPDATE Remark: This inequality does not hold in general but I am interested in the additional necessary restrictions on the $x_i$'s for the inequality to hold.

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I think it's wrong.

Try, $n=5$, $x_1=1$ and $x_2=x_3=x_4=x_5=0$.

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    $\begingroup$ Yes, I have observed that it does not generally hold. But my question is when does it hold?, i.e., what restrictions should we add to $x_i$'s for the inequality to necessary and/or sufficiently hold? $\endgroup$ – venrey Oct 22 '17 at 7:45

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