algebra precalculus - IMO 2006, A4 Problem, Idea behind the proof

IMO 2006, Problem A4, page 13:

Prove the inequality:
$$

\sum_{i \frac{n}{2(a_{1} + a_{2} + \cdots + a_{n}) }\sum_{i$$
for positive real numbers $a_{1}$, $a_{2}$, ..., $a_{n}$.

Solution (when I try to hide it, I've got the formlas broken - sorry)

Let $S = \sum\limits_{i} a_{i}$. Denote by $L$ and $R$ the expressions on the
left and right hand side of the proposed inequality. We transform $L$

and $R$ using the identity:
$$
\sum_{i < j} (a_{i} + a_{j}) = (n-1) \sum_{i}a_{i}.
$$
And thus:
$$
L = \sum\limits_{i \frac{1}{4}\sum_{i < j} \frac{ (a_{i} - a_{j})^2}{a_{i} + a_{j}}.
$$
To represent $R$ we express the sum $\sum\limits_{i
$$
\sum_{i$$
$$
\sum_{i- \frac{1}{2}\sum_{i$$
Multiplying the first of these equalities by n-1 and adding the
second one we obtain
$$

n\sum_{i \frac{1}{2}\sum_{i$$
Hence
$$
R = \frac{n}{2S} \sum_{i\frac{n-1}{4}S - \frac{1}{4}\sum_{i$$
Now compare the last two equalities. Since $S \ge a_{i} + a_{j}$
for any $i

My question: what is the idea behind this proof, and how to think it out?

Answer

The idea behind this proof: try to rewrite our inequality in the following form: $$\sum_{1\leq i

This method names SOS (Sum Of Squares). It's a very useful method.

See here:

https://math.stackexchange.com/tags/sos/info

https://math.stackexchange.com/questions/tagged/sos