I want to prove the following inequality:
$$2 \sum_{i=1}^{n-1} \sum_{j=i+1}^{n} x_{i}x_{j} \leq (n-1) \sum_{i=1}^{n} x_{i}^2 $$
I'm not sure if the notation is correct, but $\sum_{i=1}^{n-1} \sum_{j=i+1}^{n} x_{i}x_{j}$ is supposed to mean the sum of all possible (n choose 2) pairs of $x_{i}$. I don't know how to go around this, but for the specific case n=2 this would be:
$2x_{1}x_{2} \leq x_{1}^2+x_{2}^2$
Which is easily provable:
$2x_{1}x_{2} - 2x_{1}x_{2} \leq x_{1}^2+x_{2}^2 - 2x_{1}x_{2}$
$0 \leq (x_{1}-x_{2})^2$
Is this of any help for generalizing it to any amount of $n$?
We need to prove that $$2\sum_{1\leq i<j\leq n}x_ix_j\leq(n-1)\sum_{i=1}^nx_i^2$$ or $$\sum_{1\leq i<j\leq n}(x_i-x_j)^2\geq0.$$