Prove that $ \sum_{i=1}^{N} a_i \leq \sqrt{N \sum_{i=1}^{N}a_i^2} $

Question

Prove that $ \sum_{i=1}^{N} a_i \leq \sqrt{N \sum_{i=1}^{N}a_i^2} $. Well i choose $u=(1,\ldots,1)$ and $v=(a_1,\ldots,a_N)$ whit $a_i$ positive and. Apply $u \circ v \leq |u||v|$. With this now i want to prove that $\sum_{i,j}^{N}\frac{\partial u}{\partial x_i}\frac{\partial v}{\partial x_j}\leq N |\nabla u||\nabla v|$. I know that I am very close , but if somebody can help me, i will aprecciate..thank you

Answer 1

So we already know that $\sum_i a_i = u\circ v$. You then have $|u| = \sqrt{u\circ u} = \sqrt{N}$. And $|v| = \sqrt{v\circ v} = \sqrt{\sum_i a_i^2}$. Thus, $\sqrt{N\sum_i a_i^2} = |u||v|$, and so $u\circ v \leq |u||v|$ gives the desired result.

Answer 2

Recall the Cauchy-Schwarz inequality for sums:

$$\left( \sum_1^N a_n b_n \right)^2 \le \left( \sum_1^N a_n^2 \right) \left( \sum_1^N b_n^2 \right)$$

Let $b_n = 1$. Then this simplification results:

$$\left( \sum_1^N a_n \right)^2 \le \left( \sum_1^N a_n^2 \right) N$$

Taking the square root of each side gives the desired result.