How to finish this proof using Cauchy-Schwarz inequality?

Question

What should I do next?

I've tried to show that $n\sum_{i=1}^n a_i >= \sum_{i=1}^n a_i + n$ but it gets me to sth like: $\sum_{i=1}^n a_i >= \frac{n}{(n-1)}$

$a_1, a_2... \in R$

Answer 1

We are using that

• $u=\left(\frac{a_1}n,\frac{a_2}n,\dots,\frac{a_n}n\right)$
• $v=\left(1,1,\dots,1\right)$

then

$$|u\cdot v|^2=\left(\frac{a_1+a_2+\dots+a_n}{n}\right)^2\le|u|^2\cdot |v|^2=\frac{a_1^2+a_2^2+\dots+a_n^2}{n^2}\cdot n$$

$$\iff\frac{a_1+a_2+\dots+a_n}{n}\le\sqrt{\frac{a_1^2+a_2^2+\dots+a_n^2}{n}}$$