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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.

Val
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    So, you already have a hint, that induction should work. Have you at least tried to prove it? – Tomas Oct 11 '15 at 13:45
  • Of course I tried. I used induction so I got than left part of inequality's the same with x(n+1) under the module, but right now looks like root of |x1 + ... xn| + nx(n+1) and stuck on it. I also used an idea of squaring but it didn't help. – Val Oct 11 '15 at 13:56
  • You could include your calculations in your question and simply ask this community a hint for your next step. – Tomas Oct 11 '15 at 14:31
  • Thanks Tomas. I do believe it would help. – Val Oct 11 '15 at 15:25
  • Actually, your last step is wrong, where you put "more instead of less". It is true (hypothesis) that $(x_1+...+x_n)^2<n(x_1^2+...+x_n^2)$, but this does not imply your any of the conclusions. You hypothesize that $(x_1+...+x_n)^2<n(x_1^2+...+x_n^2)$ is true. But you have to prove, that from this assumption follows $(x_1+...+x_n+x_{n+1})^2<n(x_1^2+...+x_{n+1}^2)$ – Tomas Oct 11 '15 at 15:40
  • Yeah, I should prove that but I couldn't. If the last step was proven by myself I will not be emerge here for a hint. The whole reason is to prove it for $x_{n+1}$ as I understand. – Val Oct 11 '15 at 15:59

1 Answers1

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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.

kobe
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