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

Let $f(x) = x^2$. Since $f''(x) = 2 > 0$, $f$ is convex. Thus \begin{align*} f\left(\frac{x_1+x_2+\cdots + x_n}{n}\right) \leq \frac{f(x_1)+f(x_2)+\cdots+f(x_n)}{n} \end{align*} In fact, considering the function $f(x) = x^{p/q}$, where $p > q>0$, one can show that \begin{align*} \left(\frac{x_1^q+x_2^q+\cdots +x_n^q}{n}\right)^{1/q} \leq \left(\frac{x_1^p+x_2^p+\cdots +x_n^p}{n}\right)^{1/p} \end{align*}

Answer 4

for sake of contradiction suppose:
$\sqrt{\frac{a_1^2+ \cdots +a_n^2}{n}}<\frac{a_1+ \cdots +a_n}{n}\\ \frac{a_1^2+ \cdots +a_n^2}{n}<\frac{(a_1+ \cdots +a_n)^2}{n^2}\\ \frac{a_1^2+ \cdots +a_n^2}{n}<\frac{a_1^2+ \cdots +a_n^2}{n^2}+2\frac{\sum_{sym}a_1a_2}{n^2}\\ \\ (n-1)(a_1^2+\cdots+a_n^2)<2\sum_{sym}a_1a_2 \quad \quad (1)$

At this point I show by induction that the last line is wrong:
conjecture:
$(n-1)(a_1^2+\cdots+a_n^2)\geq2\sum_{sym}a_1a_2 \quad \forall \:n\geq2, n\in \mathrm{Z^+}\\ \text{let} \quad n=2\\ (a_1-a_2)^2\geq0\\ a_1^2+a_2^2-2a_1a_2>0\\ \Rightarrow a_1^2+a_2^2\geq 2a_1a_2$
Now assume for $n=k$
$(k-1)(a_1^2+\cdots +a_n^2)\geq 2\sum_{sym}a_1a_2 \quad \quad (2)\\ \text{I have from the case $n=2$}\\ a_{k+1}^2+a_1^2\geq 2a_{k+1}a_1\\ \Rightarrow 2a_{k+1}^2+a_1^2+a_2^2\geq2a_{k+1}a_1+a_{k+1}^2+a_2^2\geq2a_{k+1}a_1+2a_{k+1}a_2\\\text{proceding as above}:\\ ka_{k+1}^2+a_1^2+\cdots+a_k^2 \geq 2 a_{k+1}(a_1+\cdots+a_k) \quad \quad (3)\\ \text{now adding (2) and (3):}\\ (k-1)(a_1^2+\cdots +a_k^2)+ka_{k+1}^2+a_1^2+\cdots +a_k^2\geq2\sum_{sym}a_1a_2+2a_{k+1}(a_1+\cdots +a_k)\\ \Rightarrow k(a_1^2+\cdots +a_{k+1}^2)\geq 2\sum_{sym}a_1a_2\\ \therefore (n-1)(a_1^2+\cdots+a_n^2)\geq2\sum_{sym}a_1a_2 \quad \forall \:n\geq2, n\in \mathrm{Z^+} \quad\text{by induction}$
This clearly contradicts (1), so I have:
$\sqrt{\frac{a_1^2+ \cdots +a_n^2}{n}}\geq\frac{a_1+ \cdots +a_n}{n}$
by contradiction.

Answer 5

This is kind of a simplistic and brute-force approach:

$$\bigg(\frac{\sum_{i = 1}^n a_i}{n}\bigg)^2 \leq \frac{\sum_{i=1}^n a_i^2}{n} \iff % \bigg(\sum_{i = 1}^n a_i\bigg)^2 \leq n \sum_{i=1}^n a_i^2$$

Now open the bracket on the left of the inequality and simplify and note that the result is equivalent to:

$$\sum_{1 \; \leq \; i \; < \; j \; \leq \; n} (a_i - a_j)^2 \geq 0$$

This is obvious.