Induction proof: $(x_{1}+x_{2}+…+x_{n})^2\leqslant n(x_{1}^2+x_{2}^2+…+x_{n}^2)$ [duplicate]

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I have a problem proving this inequality but I can't get anywhere after the inductive step.

$$(x_{1}+x_{2}+...+x_{n})^2\leqslant n(x_{1}^2+x_{2}^2+...+x_{n}^2)$$ Maybe some hints would help.

marked as duplicate by clark, Jendrik Stelzner, Shailesh, user296602, yoknapatawphaJan 24 '16 at 0:11

If one wishes to proceed with a direct induction proof, then one can write

\begin{align} \left(\sum_{i=1}^{n+1}x_i\right)^2&=\left(\sum_{i=1}^{n}x_i\right)^2+2x_{n+1}\sum_{i=1}^n x_i+x_{n+1}^2\\\\ &\le n\sum_{i=1}^nx_i^2+2x_{n+1}\sum_{i=1}^nx_i+x_{n+1}^2\\\\ &=(n+1)\sum_{i=1}^{n+1}x_i^2\\\\ &-nx_{n+1}^2+2x_{n+1}\sum_{i=1}^nx_i-\sum_{i=1}^nx_i^2\\\\ &=(n+1)\sum_{i=1}^{n+1}x_i^2\\\\ &-\left(\sqrt{n}x_{n+1}-\frac1{\sqrt n}\sum_{i=1}^nx_i\right)^2\\\\ &-\left(\sum_{i=1}^nx_i^2-\frac1n \left(\sum_{i=1}^nx_i\right)^2\right)\\\\&\le (n+1)\sum_{i=1}^{n+1}x_i^2 \end{align}

And we're done!

• Why was this down voted???? – Mark Viola Jan 30 '16 at 17:19

Hint: Use the inequality of Cauchy-Schwarz.

• I once saw a proof of this without using any theorems, just by plain manipulation of inequality. Would it be possible to approach it that way? – Peter Kvačkay Jan 23 '16 at 22:10
• @PeterKvačkaj Sure, but using C-S it's a line or two :) – Anthony Peter Jan 23 '16 at 22:24

You have $S_1 \leq 1\cdot T_1, S_k = (x_1+\cdots + x_k)^2, T_k = x_1^2+\cdots +x_k^2 \Rightarrow S_n\leq nT_n\Rightarrow S_{n+1} =(x_1+\cdots + x_n + x_{n+1})^2= S_n + 2(x_1+\cdots + x_n)x_{n+1}+x_{n+1}^2 = S_n + (2x_1x_{n+1}+\cdots +2x_nx_{n+1})+x_{n+1}^2\leq S_n + (x_1^2+x_{n+1}^2)+\cdots + (x_n^2+x_{n+1}^2)+x_{n+1}^2=S_n + T_n+(n+1)x_{n+1}^2\leq nT_n+T_n+(n+1)x_{n+1}^2=(n+1)T_n+(n+1)x_{n+1}^2 = (n+1)(T_n+x_{n+1}^2)=(n+1)T_{n+1}$. Thus it is true for $n+1$ and therefore is true for all $n \geq 1$.

• In case the inequality $2ab \le a^2 + b^2$ is not obvious to OP, it follows from the fact that $0 \le (a-b)^2$. – Browning Jan 23 '16 at 22:25