Prove that
$(x_1+\dots+x_n)^2 \leq n(x_1^2 + \dots+x_n^2)$
for all positive integers n and all real numbers $x_1,….,x_n$
I am attempting a proof by induction but wasn't sure if i need the Cauchy-Schwarz Inequality or perhaps another way other than induction to prove this.
Proof
$n=1$ true
assume true for $n=k$
Now for $n= k+1$
$(x_1 + \dots +x_k + x_{k+1})^2 \leq \dots$
Best Answer
By Jensen's inequality, since $f(x)=x^2$ is convex when $x\ge 0$ we have $$ (\frac{x_1+...+x_n}{n})^2 \le \frac{x_1^2+...+x_n^2}{n} $$
whenever $x_i \ge 0$ all $i$. The case where $x_i \in \Bbb R$ follows trivially.