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.

From what you've done, $ab+bc+ca = \dfrac{(a+b+c)^2 - (a^2+b^2+c^2)}{2} \geq \dfrac{0 - 1}{2} = \dfrac{-1}{2}$, and this is the minimum value you sought. The minimum occurs when $a+b+c = 0, a^2+b^2+c^2 = 1$. To solve for $a,b,c$ you only need to find one solution of the system of $2$ equations above, then you are done. Since there are $3$ variables and only $2$ equations, you can take $c = \dfrac{1}{2}, \Rightarrow a, b$ are the solutions of the equation: $4x^2+2x-1 = 0$, thus $a = x = \dfrac{-1+\sqrt{5}}{4}, \Rightarrow b = -\dfrac{1}{2} - a = \dfrac{-1-\sqrt{5}}{4}$ by Viete's theorem on quadratic equation.
