How to prove $a^2 + b^2 + c^2 \ge ab + bc + ca$?

How can the following inequation be proven?

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

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"How can it be proved" - not solved. There is nothing to solve here . – mixedmath Sep 15 '11 at 20:00

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 $$\| x+ty \| = \left< x+ty,x+ty \right> = \|x\| + 2 \left< x,y \right>t +\|y\|t^2.$$ We know, this being a square, is greater or equal to zero. Therefore, the discriminant of the polynomial in $t$ is less or equal to zero. Which is $\left< x,y \right>^2 - \|x\|\|y\| \le 0.$ 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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beautiful mathematics – LoveFood Mar 11 '14 at 20:08