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I met this inequality :

a_i (i from 1 to n)are positive integers and not equal to each other .

Prove : $\displaystyle \frac{a_1^2+a_2^2+\cdots+a_n^2}{a_1+a_2+\cdots+a_n}\geqslant \frac{2n+1}{3}$

I tried :

$\displaystyle \Leftrightarrow \frac{n(n+1)}{2}\sum_{i=1}^n a_i^2\geqslant\frac{n(n+1)(2n+1)}{6}\sum_{i=1}^n a_i$

$\displaystyle \Leftrightarrow\sum_{i=1}^n i\sum_{i=1}^na_i^2\geqslant\sum_{i=1}^ni^2\sum_{i=1}^na_i$

It seems right.But i don't know how to prove it.Who can help me. Thanks!

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  • $\begingroup$ Cauchy-Schwarz gives a lower bound of $(n+1)/2$ for the L.H.S, but of course it is not tight here; Jensen's inequality might help. $\endgroup$ – Aravind Sep 20 '15 at 6:55
  • $\begingroup$ Yeah,you are right ,It's easy to got $\frac{n+1}{2}$. $\endgroup$ – smallsmallice Sep 20 '15 at 7:41
  • $\begingroup$ I have a feeling that using the error terms in Cauchy-Schwarz will solve it. $\endgroup$ – karvens Sep 21 '15 at 1:11
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Rewrite the inequality as $$ \sum_{i=1}^n \left(a_i^2 - \dfrac{2n+1}{3} a_i\right) \ge 0$$ Completing the square, it becomes $$ \sum_{i=1}^n \left(a_i - \dfrac{2n+1}{6}\right)^2 \ge \dfrac{n (2n+1)^2}{36}$$ Now it is clear that the least possible value of the left side (for distinct positive integers $a_i$) is when $a_i = i$ for $i=1\ldots n$, in which case the left side turns out to be $$ \sum_{i=1}^n \left( i - \dfrac{2n+1}{6}\right)^2 = \dfrac{n(2n+1)^2}{36}$$

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  • $\begingroup$ (+1) although as usual, the part "it is clear..." seems a bit painful and need a little more detail. $\endgroup$ – Quang Hoang Sep 21 '15 at 1:55
  • $\begingroup$ If $x$ and $y$ are positive integers with $x \le n < y$, then $(x - (2n+1)/6)^2 < (y - (2n+1)/6)^2$. It follows that if the $a_i$ are not the integers $1, \ldots, n$, we can reduce $\sum_i (a_i - (2n+1)/6)^2$ by replacing some $a_i > n$ by a missing member of $\{1,\ldots, n\}$. $\endgroup$ – Robert Israel Sep 21 '15 at 4:08
  • $\begingroup$ @RobertIsrael l think the proof missed something...For eg. $(1-3)^2+(2-3)^2+(3-3)^2=5$,But$(2-3)^2+(3-3)^2+(4-3)^2=1$. So the value might related to the value of$\frac{2n+1}{6}$.the proof need more details to explain this . But I think your method is right. $\endgroup$ – smallsmallice Sep 21 '15 at 10:07
  • $\begingroup$ Of course it "relates to the value of $(2n+1)/6$". It depends on the fact that $(2n+1)/6 - 1 < (n+1) - (2n+1)/6$. $\endgroup$ – Robert Israel Sep 21 '15 at 15:34

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