How to minimize $ab + bc + ca$ given $a^2 + b^2 + c^2 = 1$? The question is to prove that $ab + bc + ca$ lies in between $-1$ and $1$, given that $a^2 + b^2 + c^2 = 1$.
I could prove the maxima by the following approach. I changed the coordinates to spherical coordinates:
$a = \cos A \\
b = \sin A \cos B \\
c = \sin A \sin B$
Using that $\cos X + \sin X$ always lies between $- \sqrt 2$ and $\sqrt 2$ I proved that $a + b + c$ lies between $- \sqrt 3$ and $\sqrt 3$.
Using $(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$ I could prove that $ab + bc + ca$ is maximum at $1$ but I can't prove the minima.

Hint: $a^2+b^2+c^2+ab+bc+ca=\frac{1}{2}((a+b)^2+(b+c)^2+(c+a)^2)\geq 0$.
