Show that the arithmetic mean is less or equal than the quadratic mean

Question

I tried to solve this for hours but no success.

Prove that the arithmetic mean is less or equal than the quadratic mean.

I am in front of this form: $$ \left(\frac{a_1 + ... + a_n} { n}\right)^2 \le \frac{a_1^2 + ... + a_n^2}{n} $$

With rewriting the inequality in other forms I had no luck.

I think maybe induction would be OK, but I have no idea, how to do it in this case. Do you know a good proof for this?

Thanks!

Answer 1

The result is immediate if all the $a_i$ are zero, so we may assume that not all $a_i$ are zero. Further, by the triangle inequality, $\lvert\frac{a_1 + \cdots + a_n}{n}\rvert \le \frac{\lvert a_1\rvert + \cdots + \lvert a_n\rvert}{n}$. Therefore $\left(\frac{a_1 + \cdots + a_n}{n}\right)^2 \le \left(\frac{\lvert a_1\rvert + \cdots + \lvert a_n\rvert}{n}\right)^2$. So we may suppose additionally that all the $a_i$ are nonnegative.

Introduce a discrete random variable $X$ on the sample space $\{a_1,\ldots, a_n\}$ by letting $P(X = a_i) = a_i/n$ for $i = 1,2,\ldots n$. Then mean of $X$ is $\mu = (a_1 + \cdots + a_n)/n$, so the variance of $X$ is $$\operatorname{var}(X) = E(X^2) - \mu^2 = \left(\frac{a_1^2 + \cdots + a_n^2}{n}\right) - \left(\frac{a_1 + \cdots + a_n}{n}\right)^2$$ Since the variance is nonnegative, we deduce that $$\left(\frac{a_1 + \cdots + a_n}{n}\right)^2 \le \frac{a_1^2 + \cdots + a_n^2}{n}$$

Answer 2

By Cauchy–Schwarz inequality, $$\left(\frac{a_1}{n}1+\frac{a_2}{n}1+\cdots+\frac{a_n}{n}1\right)^2\le\left(\frac{a_1^2}{n^2}+\frac{a_2^2}{n^2}+\cdots+\frac{a_n^2}{n^2}\right)\left(1^2+1^2+\cdots+1^2\right)$$

Answer 3

Maybe a statistical proof? We know $$Var(a) = \left( \frac{\sum_i a_i^2}{n}\right)^{\frac12} - \left( \frac{ \sum_i a_i}{n}\right),$$ but we also know by definition that $Var(a) \geq 0$, therefore we know that RMS$ \geq$ average.