0
$\begingroup$

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 $n= k+1$

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

$\endgroup$
2
  • 1
    $\begingroup$ $(x_1+...+x_k+x_{k+1})^2 \le (k+1)(x_1^2+...+x_{k+1}^2)$ - $(nx_{k+1}^2-2x_{k+1}(x_1+...+x_k)+(x_1^1+...+x_k^2))$, second term is always positive. $\endgroup$ – Abstraction Nov 12 '15 at 7:40
  • $\begingroup$ It is basic RMS-AM inequality. $\endgroup$ – user249332 Nov 12 '15 at 8:51
5
$\begingroup$

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.

$\endgroup$
6
$\begingroup$

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.

$\endgroup$
1
  • 2
    $\begingroup$ Actually $f$ is convex on the whole real line, I shouldn't have bother to restrict it $\Bbb R^+$ :P $\endgroup$ – BigbearZzz Nov 12 '15 at 7:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.