Using Cauchy-Schwarz inequality to prove that the mean of n real numbers is less than or equal to the root-mean-square of those numbers Expressed mathematically, the question is to prove the that $\frac{1}{n}$ $\sum_{i=1}^{i=n}{a_i}\leqslant$ $\sqrt{\frac{1}{n}\sum_{i=1}^n{x_i}^2}.$
First of all, what form of Cauchy-Schwarz should I use or does it not matter? And is it actually Cauchy's inequality not C-S? 
My attempt, using the vector form:
$|u\cdot v| \leq |u||v|$ where $u,v \in R^n$
$(\sum_{i=1}^{n}{u_iv_i})^2 < \sum_{i=1}^{n}{(u_i)}^2\sum_{i=1}^{n}{(v_i)}^2$ 
I know that from this step on I should find some vector v and eventually take the square root of the expression? 
 A: The Cauchy Schwarz inequality is
$$\left(\sum_{i=1}^{n}{u_iv_i}\right)^2 \le \sum_{i=1}^{n}{u_i}^2\sum_{i=1}^{n}{v_i}^2$$
so let
$$u_i=\frac 1{\sqrt n}\qquad;\qquad v_i=\frac{x_i}{\sqrt n}$$
then
$$\frac1 {n^2}\left(\sum_{i=1}^{n}x_i\right)^2 \le\frac1 n\sum_{i=1}^{n}{x_i}^2 $$
and take the square root.
A: In $(\sum_{i=1}^{n}{u_iv_i})^2 < \sum_{i=1}^{n}{(u_i)}^2\sum_{i=1}^{n}{(v_i)}^2$,
take $u_i=x_i$ and $v_i=1$, for each $i=1(1)n$. You will get $(\sum_{i=1}^{n}{x_i})^2 < \sum_{i=1}^{n}{(x_i)}^2.\sum_{i=1}^{n}{(1)}^2$, take squareroot and divide by $n$.
A: I imagine that what you need to show is
$$
\frac{1}{n}\sum_{i=1}^{i=n}{a_i}\leqslant\sqrt{\frac{1}{n}\sum_{i=1}^n{a_i}^2},
$$
which is equivalent to showing that
$$
\sum_{i=1}^{i=n}{a_i}\leqslant\sqrt{n\sum_{i=1}^n{a_i}^2},
$$
which is Cauchy inequality for $b_1=b_2=\cdots=b_n=1$.
Indeed
$$
\sum_{i=1}^{i=n}{a_i}=\sum_{i=1}^{i=n}{a_i}b_i\leqslant\sqrt{\sum_{i=1}^n{a_i}^2\sum_{i=1}^n{b_i}^2}=
\sqrt{n\sum_{i=1}^n{a_i}^2}.
$$
A: $\frac{1}{n}$ $\sum_{i=1}^{i=n}{a_i}\leqslant$ $\sqrt{\frac{1}{n}\sum_{i=1}^n{a_i}^2}$ is same as $|avg(a)|\leqslant rms(a).$
So, I proved $|avg(a)|\leqslant rms(a).$  
Cauchy-Schwarz inequality tells-
$$|a^Tb|\leqslant||a||||b||$$
Let, $b$ is one(1) vector.  
Then,
 $$|a^Tb|\leqslant||a||||b||$$  
$$|n\ avg(a)|\leqslant||a||,\ \text{as } ||b||=1$$  
$$|avg(a)|\leqslant\frac{||a||}{n}$$  
$$|avg(a)|\leqslant\frac{||a||}{\sqrt{n}}$$  
$$|avg(a)|\leqslant rms(a)$$
