# Show that $a^2 + b^2 + c^2 \geq ab + bc + ca$ for all positive integers $a$, $b$, and $c$ [duplicate]

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Show that $a^2 + b^2 + c^2 \geq ab + bc + ca$ for all positive integers $a$, $b$, and $c$.

I am not sure how to approach this problem. Should I divide this problem into multiple cases based on whether $a$,$b$,$c$ and odd/even or is there a more general solution?

## marked as duplicate by Martin Sleziak, Macavity, Claude Leibovici, Najib Idrissi, Davide GiraudoJan 30 '15 at 9:18

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• it's a little trick called completing the square: expand $(a-b)^2+(b-c)^2+(c-a)^2\ge 0$ – Myself Jan 29 '15 at 22:41
• This is a special case of the so called rearrangement inequality. – Arthur Jan 29 '15 at 22:42
• This holds for all reals, not just positive integers. – Macavity Jan 30 '15 at 1:23

## 2 Answers

$0\le\dfrac12\left((a-b)^2+(b-c)^2+(c-a)^2\right)=\dfrac12\left(a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2\right)=a^2+b^2+c^2-ab-bc-ca\implies a^2+b^2+c^2\ge ab+bc+ca$

• Thank you. The book goes over basic problem techniques and then drops this mind bender with no mention of completing squares or rearrangement inequalities. Is this the only way to solve this problem? – user3699546 Jan 29 '15 at 22:47
• @user3699546. There may be more ways, but this one is the simpliest. You can prove any polynomial inequality reducing to sum of squares. Also, please consider accepting my answer if you found it to be useful. – user164524 Jan 30 '15 at 0:16
• @Mathematician171 Just to be accurate, note there are polynomial inequalities which do not reduce to sum of squares. Eg $x^2y^2+y^2z^2+z^2x^2+1\ge 4xyz$ which is readily verified by AM-GM. – Macavity Jan 30 '15 at 2:59
• @Macavity. Maybe it is easier to prove that $x^2y^2+y^2z^2+z^2x^2+1\ge4xyz$ using AM-GM, but it doesn't mean that this inequality cannot be proven reducing to sum of squares. Note that proof of AM-GM is based on reducing to sum of squares. – user164524 Jan 30 '15 at 9:44
• @Mathematician171 Well, I would like to see you try for that polynomial ;). Do you have a proof or reference to the statement you made that "any polynomial inequality" can be proven by reducing to sum of squares? The e.g. I gave is a positive semi-definite polynomial which can be proven not to have any sum of squares representation in $\mathbb R[x, y]$. There are others, e.g. search for "Motzkin polynomial". – Macavity Jan 30 '15 at 9:47

Another way:

$$a^2 + b^2 + c^2 = \frac{a^2 + b^2}{2} + \frac{b^2 + c^2}{2} + \frac{c^2 + a^2}{2} \ge ab + bc + ca$$

The last part follows from the fact that $x^2 + y^2 \ge 2xy$ for all $x, y$ (you can get this by observing $x^2 + y^2 - 2xy = (x - y)^2 \ge 0$ for all $x,y$).