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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?

Ab urbe condita
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user3699546
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2 Answers2

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

  • 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. –  Jan 30 '15 at 00:16
  • @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 02:59
  • @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. –  Jan 30 '15 at 09:44
  • @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 09:47
  • @Macavity. This is my solution. Tell me if I made any mistake. –  Feb 03 '15 at 18:36
  • @Mathematician171 I stopped at the second line where you seem to imply $4^4=16$;) More seriously, you said that any polynomial inequality reduces to a sum of squares. From your attempt it seems you mean, it could be after some manipulations, shown equivalent to the statement that squares of some arbitrary functions are positive. Now that is probably true, given everything can be reduced to finally the definition of $\ge$ sign. Directly restating an inequality (as you did in the post) as the sum of squares of polynomials, is a different thing altogether. – Macavity Feb 04 '15 at 08:37
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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 $x,y$).

kobe
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