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How can the following inequation be proven?

$$a^2 + b^2 + c^2 \ge ab + bc + ca$$

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  • 5
    $\begingroup$ "How can it be proved" - not solved. There is nothing to solve here . $\endgroup$ – davidlowryduda Sep 15 '11 at 20:00
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    $\begingroup$ I think $a,b,c$ should be greater or equal with $0$. $\endgroup$ – Iuli Sep 6 '17 at 11:28
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Try $(a-b)^2+(b-c)^2+(c-a)^2 \ge0$

Compute lhs, divide by two and rearrange.

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This is a specific form of Cauchy-Schwarz inequality.

Let $x = (a, b, c)$ and $y = (b, c, a)$ as vectors.

The inequality is $ | \left< x,y \right>| \le \|x\|\|y\|. $ with standard inner product definition. One neat trick to prove this is using an auxilary parameter $t,$ and expanding $$ \| x+ty \|^2 = \left< x+ty,x+ty \right> = \|x\|^2 + 2 \left< x,y \right>t +\|y\|^2t^2.$$ We know, this being a square, is non-negative. Therefore, the discriminant of the polynomial in $t$ is less or equal to zero. Which is $\left< x,y \right>^2 - (\|x\|\|y\|)^2 \le 0.$ Substituting the values for $x$ and $y$ will do the job.

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This is also a consequence of the Rearrangement inequality.

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  • $\begingroup$ beautiful mathematics $\endgroup$ – LoveFood Mar 11 '14 at 20:08
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$$\sum_{cyc}(a^2-ab)=\frac{1}{2}\sum_{cyc}(a^2-2ab+b^2)=\frac{1}{2}\sum_{cyc}(a-b)^2\geq0$$

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From Cauchy-Schwarz

$ab+bc+ac=\sqrt{a^2}\sqrt{b^2}+\sqrt{b^2}\sqrt{c^2}+\sqrt{a^2}\sqrt{c^2} \leq \sqrt{a^2+b^2+c^2}\sqrt{a^2+b^2+c^2}$

Moving on;

$ab+bc+ac \leq a^2+b^2+c^2$

Done!

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İf $\ c> a ,a^2+c^2 \gt 2ac $because$\ (a-c)^2 \gt 0$ İf $\ c>b>a ,c^2+b^2/2+a^2/2 \gt ac+bc $ and $\ b^2/2+a^2/2 \gt ab $ sum of them $\ a^2+b^2+c^2 \gt ab+bc+ac $ İf $\ c=b \gt a $ or $\ a=b=c $ it can be solved with same logic.

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