# Prove $(x_1+…+x_n)^2 \leq n(x_1^2 + …+x_n^2)$

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 $n= k+1$

$(x_1 + \dots +x_k + x_{k+1})^2 \leq \dots$

• $(x_1+...+x_k+x_{k+1})^2 \le (k+1)(x_1^2+...+x_{k+1}^2)$ - $(nx_{k+1}^2-2x_{k+1}(x_1+...+x_k)+(x_1^1+...+x_k^2))$, second term is always positive. – Abstraction Nov 12 '15 at 7:40
• It is basic RMS-AM inequality. – user249332 Nov 12 '15 at 8:51

## 2 Answers

HINT: Use Cauchy-Schwarz inequality- $$(1^2+1^2+1^2+...+1^2)(x_1^2+x_2^2+x_3^2+...+x_n^2) \ge (x_1+x_2+x_3+...+x_n)^2$$ and manouvre accordingly.

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.

• Actually $f$ is convex on the whole real line, I shouldn't have bother to restrict it $\Bbb R^+$ :P – BigbearZzz Nov 12 '15 at 7:44