I need to show that:
$$ {\sum\limits_{i=1}^n {|x|} } \leq \sqrt{n\sum\limits_{i=1}^n |x|^2 } $$
I tried to square both sides so I would get:
$$ \left({\sum\limits_{i=1}^n {|x|} }\right)^2 = \left(\sum_{i=1}^{N}|x_i|^2+2*\sum_{i,j,i j}|x_i||x_j|\right) \leq n\sum\limits_{i=1}^n |x|^2 $$
but it just doesn't seem to work...
I know that on both sides we have $n^2$ elements, I just don't know how to compare them.
Make sure you don't have a typo, and that you copied the question correctly. I suspect you need to work with the following:
$$\left(\sum_{i=1}^n|x_i|\right)^2 \le n \sum_{i=1}^n|x_i|^2.$$
Now, each side of the inequality has $n^2$ terms.
You can use the Cauchy-Schwarz Inequality. As applied to Euclidean space $\mathbb{R}^n$:
$$\left(\sum_{i=1}^n x_iy_i\right)^2 \le \left(\sum_{i=1}^nx_i^2 \right)\left(\sum_{i=1}^n y_i^2 \right) $$
For your problem, $$\left(\sum_{i=1}^n x_iy_i\right)^2 = \left(\sum_{i=1}^n x_i\cdot 1\right)^2$$
Alternatively, if you have the following inequality to prove: $$\left(\sum_{i=1}^n|x_i|\right) \le \sqrt{n \sum_{i=1}^n|x_i|^2}.$$
Then simply square both sides of this inequality to obtain the inequality at the top, and proceed as suggested.
Because $x^2$ is a convex function, Jensen's Inequality says $$ \left(\frac1n\sum_{k=1}^nx_k\right)^2\le\frac1n\sum_{k=1}^nx_k^2 $$ Multiplying by $n^2$ yields $$ \left(\sum_{k=1}^nx_k\right)^2\le n\sum_{k=1}^nx_k^2 $$ Taking the square root gives $$ \sum_{k=1}^nx_k\le\sqrt{n\sum_{k=1}^nx_k^2} $$
