I am having a little difficulty trying to solve a beginner proof in the topic of inner product spaces.
The statement says,
suppose $a_1,…,a_n \in \mathbb R$,
Prove that $(a_1+…+a_n)^{2}/n \le a_1^2+…+a_n^2$
I am thinking I definitely need to use the Cauchy shwarz inequality but there is a few things I am unclear on. Is $(a_1+…+a_n)^{2}$= $\lt a_1+…+a_n,a_1…+a_n \gt $, and how else to approach. Any answers or help would be appreciated, thanks a lot!
On
If we're using the form of Cauchy-Schwarz that says $\lvert\langle x, y\rangle\rvert^2 \leq \langle x, x \rangle \langle y, y\rangle$, we can expand this component-wise. Then we have
$$(x_1y_1 + x_2y_2 + \ldots x_ny_n)^2 \leq (x_1^2 + x_2^2 + \ldots + x_n^2)(y_1^2 + y_2^2 + \ldots +y_n^2).$$
Your inequality looks like
$$\frac{(a_1 + a_2 + \ldots a_n)^2}{n} \leq (a_1^2 + a_2^2 + \ldots a_n^2).$$
If you multiply both sides by $n$, it seems likely that you want to use $x = (a_1, a_2, \ldots, a_n)$. What would be a suitable $y$?
Hint: $$ \frac{(a_1 + \cdots + a_n)^2}{n} = \left( \frac{a_1}{\sqrt{n}} + \cdots + \frac{a_n}{\sqrt{n}} \right)^2. $$ Can you use Cauchy-Schwarz on this?