# Inequality question: Sum of squares and determining minimum value.

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.

-
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 $-3$.
@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