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Given that $a^2 + b^2 + c^2 = 6\;$, determine the minimum value of $ab + bc + ca\;$.

I know that $a^2 + b^2 + c^2 ≥ ab + bc + ca\;$, so that means $6 ≥ ab + bc + ca\;$, but I don't know where to go from there.

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Are you familiar with Lagrange multipliers? – Alex Becker Jul 23 '12 at 21:46
Sorry, I am not familiar with that. – saiBlue Jul 23 '12 at 21:54

You need a lower bound for $ab+bc+ca$. The inequality $ab+bc+ca \le a^2+b^2+c^2$ goes in the "wrong" direction. It is useful for producing an upper bound for $ab+bc+ca$.

Use instead the fact that $$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)=6+2(ab+bc+ca).$$

So to minimize $ab+bc+ca$ we must minimize $(a+b+c)^2$. Clearly $(a+b+c)^2 \ge 0$ always. But we can arrange for $a+b+c$ to be $0$ in infinitely many ways. One example is $a=\sqrt{3}$, $b=-\sqrt{3}$, $c=0$. (Any point where the plane $x+y+z=0$ meets the sphere $x^2+y^2+z^2=6$ will do it.)

So the minimum value of $ab+bc+ca$ under our constraint is $-3$.

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Ok thanks, that helps a lot. :D So to clarify, the actual values of a, b and c are irrelevant, as long as their sum is 0? – saiBlue Jul 23 '12 at 22:45
@saiBlue: And of course $a^2+b^2+c^2=6$. I made the side comment about the sphere to add some geometric intuition. There is a full circle of points at which $ab+bc+ca$ is minimized. – André Nicolas Jul 23 '12 at 22:48
Oh ok, thanks again for the help. :) – saiBlue Jul 23 '12 at 22:56

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