If $a^2+b^2+c^2=1$ then prove the following. If $a^2+b^2+c^2=1$, prove that $\frac{-1}{2}\le\ ab+bc+ca\le 1$.
I was able to prove that $ ab+bc+ca\le 1$. But I am unable to gain an equation to prove that
$ \frac{-1}{2}\le\ ab+bc+ca$ .
Thanks in advance !

(a - b)^2 + (b - c)^2 + (a - c)^2 = 2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ac = 2 - 2(ab + bc + ac). This is how we prove that ab + bc + ca <= 1.
Now: 
If ab + bc + ac < -1/2, then 2(ab + bc + ac) < -1.
(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2(a + b)(c) = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc = 1 + 2(ab + ac + bc) < 1 + (-1) = 0. Which is impossible.
(Edited).
