How to show $x_k \in \mathbb R, \frac {\sum{x_k}}{n} \leq \left(\frac{\sum{x_k}^2}{n}\right)^n$?

Question

Prove that, for arbitrary real numbers $x_1,x_2,x_3...,x_n$

$$\frac{x_1+x_2+x_3...+x_n}{n} \leq \left(\frac{x_1^2+x_2^2+x_3^2...+x_n^2}{n}\right)^n$$

What theorem would you use to prove the following inequality? I would also like to know how to learn more about inequalities.Thanks

It's false. Letting $x_1=x_2=\cdots=x_n=x$, then it becomes $x\le x^{2n}$, which implies $x\ge 1$ or $x\le 0$ — 23rd, Jun 14 '13 at 12:50
it is no very complicated. I would like to offer an answer but I have no time. Surely you will find answers for your questions in this document: Inequality — Iuli, Jun 14 '13 at 12:41

score 1 · Answer 1 · answered Jun 19 '13 at 01:47

Take $n=1$, $x_1=0.5$. (Fails with each $n$)

$$0.5=\frac{1}{n}\sum x_k>\left(\frac{1}{n}\sum x_k^2\right)^n=0.25$$

Probably you want to prove

$$\text{AM}=\frac{1}{n}\sum x_k\leq \sqrt{\frac{1}{n}\sum x_k^2}=\text{RMS}$$

which is true, and for $n=2$ the proof from Proofs Without Words

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1 Answers1