Minimise $ab+bc+ac$ Let $a,b,c \in \mathbb R$, and $a^2+b^2+c^2=1$
How can I calculate the minimum value of $ab+bc+ac$? (i.e. most negative)
I've tried using the fact that $(a-b)^2+(b-c)^2+(a-c)^2 \ge 0$ but this gives an inequality in the wrong direction.

Note that $f(a,b,c) = ab+bc+ac = {1 \over 2} ((a+b+c)^2-(a^2+b^2+c^2))$.
Hence $\min \{ f(a,b,c) | a^2+b^2+c^2=1 \} = {1 \over 2} \min \{ (a+b+c)^2-1 | a^2+b^2+c^2=1  \}$.
Note: See Barry's comment below for a simpler alternative to the following.
One way that doesn't use Lagrange multipliers is to change basis.
Let $u_1 = {1 \over \sqrt{3}} (1,1,1)$ and let $u_2,u_3$ be such that $u_k$ form an orthonormal basis. In this basis the problem becomes
${1 \over 2}\min \{ 3x^2-1 | x^2+y^2+z^2 = 1\} $ and, by inspection, we see that
the minimum is $-{1 \over 2}$.
