<s>[INST] How to show that if $a^2+b^2+c^2=1$ then $\frac{-1}{2}\le ab+ac+bc \le 1$ How to show that if $a^2+b^2+c^2=1$ then:
$$\frac{-1}{2}\le ab+ac+bc \le 1$$

From the assumption I see that:
$$-\frac{1}{2}\le\frac{\left(a+b+c\right)^{2}-1}{2}=\frac{\left(a+b+c\right)^{2}-\left(a^{2}+b^{2}+c^{2}\right)}{2}=ab+ac+bc $$
On the other hand $a^2=1-(b^2+c^2)<1$,then same does hold for $b^2$ and $c^2$ which shows that $-1<a<1$ and so :$$-3<a+b+c<3$$ $$\implies ab+ac+bc =\frac{\left(a+b+c\right)^{2}-1}{2}<4$$
But it's does not what I've been asked to prove,so how can I show that?
 [/INST] Hint:
For the 2 inequalities you want to prove, expand the left hand sides of:
\begin{align}
(a+b+c)^2&\geq 0,\\
(a-b)^2+(b-c)^2+(c-a)^2&\geq 0.
\end{align}
And remember to use $a^2+b^2+c^2=1$.
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