How can I prove this inequality using Cauchy-Schwarz inequality

Question

I am reading introduction to functional analysis with application of Kreyzig. The question in excercise is "Show that Cauchy-Schwarz inequality implies that $$(|a_{1}|+|a_{2}|+...+|a_{n}|)^2\leq n(|a_{1}|^2+|a_{2}|^2+...+|a_{n}|^2).$$" The Cauchy-Schwarz inequality is $$\sum|\zeta_{j}\eta_{j}|\leq (\sum|\zeta_{j}|^p)^{\frac{1}{p}} (\sum|\eta_{j}|^q)^{\frac{1}{q}}$$ Thanks in advance

Answer 1

Due to Cauchy Schwartz, $[\sum_{i=1}^{n} {|a_i|}^2][\sum_{i=1}^{n} {|b_i|}^2\geq \sum_{i=1}^{n} {(|a_i||b_i|)}^2$

put $b_i=1$ for all i you get your answer.

Answer 2

Hint: Observe we may suppose $a_i\ge 0$ for all $i=11,\dots n$. Next apply Cauchy-Schwarz to the vectors $\;(a_1,a_2, \dots, a_n)\;$ and $\;(1,1,\dots, 1)$.