<s>[INST] module of sum is less than... Somehow need to prove:
$$|x_1 + x_2 + ... + x_n| \le \sqrt{n(x_1^2 + x_2 ^ 2 + ... +x_n^2)}$$
        $x_i$ is a real number; $i = 1,...,n$
Here's mentioned that mathematical induction should help. So I tried to use it but right after squaring, cause I had no idea how to do it without squaring first. 
So it looks like: 
$$(x_1 + x_2 + ... + x_n + x_{n+1})^2 \le (n+1)(x_1^2 + x_2^2 + ... +x_n^2 + x_{n+1}^2) $$ 
And some transformation based on the induction hypothesis(part which is more should be transformed into less:
$$(x_1 + x_2 + ... + x_n + x_{n+1})^2 \le (x_1 + x_2 + ... + x_n )^2  + (n+1)x_{n+1}^2$$
Community I need a hint for the next step or even another idea for making a proof.
 [/INST] Your induction hypothesis is $|x_1 + \cdots + x_n| \le \sqrt{n(x_1^2 + \cdots + x_n^2)}$, which becomes $(x_1 + \cdots + x_n)^2 \le n(x_1^2 + \cdots + x_n^2)$ by squaring. Write
\begin{align}(x_1 + \cdots + x_n + x_{n+1})^2 &= (x_1 + \cdots + x_n)^2 + x_{n+1}^2 + 2(x_1 + \cdots + x_n)x_{n+1}\\
&= (x_1 + \cdots + x_n)^2 + x_{n+1}^2 + 2x_1x_{n+1} + \cdots + 2x_nx_{n+1}.\tag{*}
\end{align}
Since $2xy \le x^2 + y^2$ for all $x,y\in \Bbb R$, then 
$$2x_1x_{n+1} + \cdots + 2x_nx_{n+1} \le (x_1^2 + x_{n+1}^2) + \cdots + (x_n^2 + x_{n+1}^2) = (x_1^2 + \cdots + x_n^2) + nx_{n+1}^2.$$
Using the induction hypothesis, we find that the expression (*) is less than or equal to 
$$n(x_1^2 + \cdots + x_n^2) + x_{n+1}^2 + (x_1^2 + \cdots + x_n^2) + nx_{n+1}^2,$$
which is 
$$(n+1)(x_1^2 + \cdots + x_n^2) + (n+1)x_{n+1}^2 = (n+1)(x_1^2 + \cdots + x_n^2 + x_{n+1}^2)$$
as desired. 
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