TITLE: a_i are positive integers and not equal to each other .prove : $\frac{a_1^2+a_2^2+\cdots+a_n^2}{a_1+a_2+\cdots+a_n}\geqslant \frac{2n+1}{3}$
QUESTION [9 upvotes]: 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!

REPLY [6 votes]: 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
$$ \sum_{i=1}^n \left( i - \dfrac{2n+1}{6}\right)^2 = \dfrac{n(2n+1)^2}{36}$$