<s>[INST] Show that the arithmetic mean is less or equal than the quadratic mean 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!
 [/INST] 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.
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