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This question already has an answer here:

$a^2 +b^2 +c^2 \ge a(b+c)+bc$, this equation is supposed to be correct, although I factorized and tried to solve prove in many ways, I cant explain well enough!

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marked as duplicate by Shailesh, David K, mrs, BLAZE, user228113 Feb 15 '16 at 7:23

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    $\begingroup$ I have marked this as a duplicate. However, the most intuitive (direct) way of explaining this perhaps lies in the fact that of all rectangles of a given perimeter, the square has the largest area. Thanks @BrianTung. Draw 3 squares of lengths a, b, c. Now construct three rectangles a,b then b,c and then c,a out of them. $\endgroup$ – Shailesh Feb 15 '16 at 6:09
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If $a-b,b-c,c-a$ are real(which will be true if a,b,c are real),

$$(a-b)^2+(b-c)^2+(c-a)^2\ge0$$

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Consider $f(a) = a^2 -a(b+c) + b^2 + c^2 - bc \Rightarrow \triangle = (b+c)^2 - 4(b^2+c^2-bc) = -3b^2-3c^2+6bc = -3(b-c)^2 \leq 0\Rightarrow f(a) \geq 0$.

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