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how can i use AM-GM inequality to show that $$x_1^2+x_2^2+\cdots+x_{2020}^2\geqslant x_1x_2+x_2x_3+\cdots+x_{2020}x_1 ?$$ Where $x_1,x_2,\ldots,x_n$ are any real numbers.

to get an idea I tried to do for particular cases with $n=3,4.$

for $n=3$, I could use AM-GM doing $$x_1^2+x_2^2+x_3^2=\frac{x_1^2+x_2^2}{2}+\frac{x_2^2+x_3^2}{2}+\frac{x_3^2+x_1^2}{2}\geqslant x_1x_2+x_2x_3+x_3x_1.$$ for $n=4$ couldn't with that same idea. but it is possible to show developing the inequality below $$(x_1-x_2)^2+(x_2-x_3)^2+(x_3-x_4)^2+(x_4-x_1)^2 \geqslant 0.$$

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  • $\begingroup$ "for $n=4$ couldn't with that same idea." Why? $\endgroup$
    – xskxzr
    Commented Jan 10, 2020 at 2:20
  • $\begingroup$ If they are real numbers they can be negative . How will you apply AM-GM if that's the case? $\endgroup$ Commented Jan 10, 2020 at 2:26
  • $\begingroup$ @ArchisWelankar AM-GM Inequality still holds if $n$ is even. We can simply replace each $x_i$ with $|x_i|$. $\endgroup$ Commented Jan 10, 2020 at 2:29
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    $\begingroup$ You might want to see this. $\endgroup$
    – ViHdzP
    Commented Jan 10, 2020 at 2:33
  • $\begingroup$ Cauchy Schwarz will do it to, if you write it as $\big \vert \mathbf x \big(P \mathbf x\big)\big \vert \leq \big \Vert \mathbf x\big \Vert \big \Vert P\mathbf x\big \Vert = \big \Vert \mathbf x\big \Vert^2$ with $P:=\begin{bmatrix} 0 & 0 &0 & \cdots &0 & 1\\ 1 & 0 &0 &\cdots &0 & 0\\ 0 & 1 &0 & \cdots &0 &0 \\ 0 & 0 &1& \cdots &0 & 0\\ 0 & 0 &0& \ddots &0 & 0\\ 0 & 0 &0& \cdots &1 & 0 \end{bmatrix}$ $\endgroup$ Commented Jan 10, 2020 at 6:12

2 Answers 2

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The idea is the same: $$ x_1x_2 + x_2x_3 + \cdots + x_{2020}x_1 \leq \frac{x_1^2 + x_2^2}{2} + \frac{x_2^2 + x_3^2}{2} + \cdots + \frac{x_{2020}^2 + x_1^2}{2} = x_1^2 + x_2^2 + \cdots + x_{2020}^2 $$

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  • $\begingroup$ Suppose for all odd i $x_i<0$ then lhs will be negative and rha which is sum of squares will be positive that's an obstacle which needs to be tackled. $\endgroup$ Commented Jan 10, 2020 at 2:30
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    $\begingroup$ @ArchisWelankar That's no issue, since $x_1x_2 + x_2x_3 + \cdots + x_{2020}x_1 \leq |x_1||x_2| + |x_2||x_3| + \cdots + |x_{2020}||x_1|$. $\endgroup$ Commented Jan 10, 2020 at 2:31
  • $\begingroup$ Yeah that will do it. $\endgroup$ Commented Jan 10, 2020 at 2:31
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Let $x_{n+1}=x_1.$ We have $$\sum_{i=1}^nx_i^2-\sum_{i=1}^nx_ix_{i+1}=$$ $$=\sum_{i=1}^n \frac {1}{2}(x_i^2+x_{i+1}^2)-\sum_{i=1}^nx_ix_{i+1}=$$ $$=\sum_{i=1}^n\frac {1}{2}(x_i^2+x_{i+1}^2-2x_ix_{i+1})=$$ $$=\sum_{i=1}^n\frac {1}{2}(x_i-x_{i+1})^2\ge 0.$$ See " Cauchy-Schwartz inequality " in Wikipedia. In the notation of that article, this Q is a special case with $u=(x_1,...,x_n)\in \Bbb R^n$ and $v=(x_2,...,x_{n+1})\in \Bbb R^n$ with $x_{n+1}=x_1. $

BTW. In $\Bbb R^m,$ with $m\ge 2,$ define the distance from $A=(a_1,...,a_m)$ to $B=(b_1,...,b_m)$ as $AB=(\sum_{i=1}^m(a_i-b_i)^2)^{1/2}.$ Now take any two vectors $A,B \in \Bbb R^m,$ neither of them equal to the origin $O,$ and neither of them a scalar (real-number) multiple of the other. Then the $2$-dimensional vector subspace generated by $A$ and $B$ is geometrically isomorphic to $\Bbb R^2,$ and $$\sum_{i=1}^ma_ib_i=AO\cdot BO\cdot \cos \angle AOB.$$

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