Prove the following:

The summation ranges from 1 to n.

${\sum{a_i^2}\ge \frac1n }$ provided $\sum a_i$=1.

without using the method used to prove chebyshev's inequality.

I can do this very easily assuming that we can arrange $a_i$ such that $a_1\ge$$a_2\ge$...$\ge a_n$. but I want to do it without using it.

I derive two conditions but cannot continue from there:

further work:

${\sum{a_i^2} }\le$$\sum a_i\sum a_i$$\implies \sum{a_i^2}\le 1$

and $n=\sum\sum a_i$.