I'm not sure if this will help you explicitly, but there's a little identity that might contribute to your solution
We know that
[tex]\displaystyle\left( \sum_i a_i \right) ^2 \geq 0[/tex]
by non-negativity of a square, with equality holding iff the summand is identically 0. Furthermore, we can expand this to
[tex]\displaystyle\left( \sum_i a_i \right) ^2 = \sum_i a_i^2 + 2 \sum_{i<j} a_i a_j[/tex]
Thus
[tex]\sum_i a_i^2 + 2 \sum_{i<j} a_i a_j \geq 0[/tex]
Also, this can probably be generalized but off the top of my head I'm not too sure how, but
[tex](x-y)^2 \geq 0[/tex]
[tex]\Rightarrow x^2+y^2-2xy \geq 0[/tex]
[tex]\Rightarrow 2xy \leq x^2+y^2[/tex]