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 !

If $ab+bc+ca < \dfrac{-1}{2} \Rightarrow 2(ab+bc+ca) + 1 < 0 \Rightarrow 2(ab+bc+ca) + a^2+b^2+c^2 < 0 \Rightarrow (a+b+c)^2 < 0$. Contradiction.
