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This question already has an answer here:

How do I prove that for any $a, b, c \in \mathbb{R}$ the inequality $a^2+b^2+c^2 \geq ab+bc+ac$ is true?

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marked as duplicate by JonMark Perry, GNUSupporter 8964民主女神 地下教會, Martin R, Namaste, Arnaud D. Feb 19 '18 at 15:26

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Or $$a^2+b^2\geq 2ab$$ $$b^2+c^2\geq 2bc$$ and $$c^2+a^2\geq 2ca$$ Now sum all these and divide by 2.

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  • $\begingroup$ Thank you ChristianF!!🙂🙂 $\endgroup$ – Baby Feb 20 '18 at 19:32
  • $\begingroup$ You are welcome. Which answer will you accept? $\endgroup$ – Maria Mazur Feb 20 '18 at 19:34
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it is $$(a-b)^2+(b-c)^2+(c-a)^2\geq 0$$

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  • $\begingroup$ I think this method is easier. Thank you so much for helping me!!! $\endgroup$ – Baby Feb 20 '18 at 19:31
  • $\begingroup$ thats nice and have a nice day too $\endgroup$ – Dr. Sonnhard Graubner Feb 20 '18 at 19:32
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Solution 1: rearragement inequality

Hint: WLOG, assume $a \le b \le c$. Use rearrangement inequality to conclude the desired inequality.

direct sum $a^2+b^2+c^2 \ge$ random sum $ = ab+bc+ca$, with equality holds if and only if $a = b = c$.

Solution 2: Cauchy-Schwartz inequality

Hint: set $v_1 = (a,b,c), v_2 = (b,c,a)$.

$ab+bc+ca = |\langle v_1, v_2 \rangle| \le ||v_1|| ||v_2|| = a^2+b^2+c^2$

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