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: 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.
A: $0\leq(a+b+c)^2=(a^2+b^2+c^2)+2(ab+ac+bc)\implies ab+ac+bc\geq-\dfrac{1}{2}$.
A: this is not a rigorous proof but a geometric argument. look at the transformation $T$ that sends $(a,b,c)^T$ to $(b,c,a)$. this is a rotation by $120^\circ$ about the axis $(1,1,1)^T$ the minimum value of the $T(a,b,c)^T.(a,b,c)$ happens when $(a,b,c)$ is perpendicular to the axis of rotation. this can be seen if you resolve $(a, b, c)^T$ into a component parallel to the axis and perpendicular to the axis.
A: (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).
A: $ab + bc + ca$ can be written as $x^T A x$ where $x = (a,b,c)$ as a column vector and the rows of $A$ are
$$.5, .5, 0 $$
$$.5, 0, .5$$ 
$$0 , .5,.5$$ 
$A$ is symmetric so we can find the extremal values on the unit disk by looking at the eigenvalues which all happen to be -1/2, 1/2 and 1 (this follows by diagonalizing A and using the eigenvector basis).
Eigenvalues computed by Wolfram Alpha
