Prove that the series is less than the square of a series. Prove that $(\sum_{i=1}^n a_i)^2 \leq n \sum_{i=1}^na_i^2$ for $a_1, ..., a_n$. Hint: You may want to use the triangle inequality or Cauchy-Shwartz inequality.
I'm trying to prove this preposition. here's what I've done so far...
$(\sum_{i=1}^n a_i)^2$ = $\sum_{i=1}^n\sum_{j=1}^n a_ia_j$
Then for each $i$, $\sum_{j=1}^n a_ia_j = \langle a_i, a\rangle$ (Dot product)
By Cauchy Shwartz, $\langle a_i, a_j\rangle \leq ||a_i||\:||a||  $
To me it looks like this is a "nested for loop" which means there will be an $n$ number of dot products producted, or $n \cdot ||a|| \: ||a||$
But I'm stuck, so please help and don't really know what direction this proof is going in.
Thanks.  

Hint:
Consider the vectors $\underbrace{(1,1,\dots,1)}_{n\;1\text{s}}$ and $\;(a_1, a_2,\dots,a_n)$.
