Vtyjkyuk

If $a,b,c$ belong to set of real numbers and $ a^2+b^2+c^2=1$ then prove that $ab+bc+ca$ belongs to $[-frac12,1]$

I have tried AM>GM>HM(progressions mean inequality) but I am unable to do anything.
I have even attempted to assume a,b and c to be sides of an triangle and using $|a-b|<|c|$ and squaring but still I don't get the right-hand limit right.

Please tell me how to prove it.

Let's denote $S = ab+bc+ca$.

For the lower bound:
$$(a+b+c)^2 geq 0 Leftrightarrow a^2+b^2+c^2 + 2S = 1 + 2S geq 0 Leftrightarrow boxedS geq -frac12$$
For the upper bound we use Cauchy-Schwarz:
$$1+2S = (a+b+c)^2 = (1cdot a + 1cdot b+1cdot c)^2 leq 3cdot (a^2+b^2+c^2) = 3 $$$$Leftrightarrow 1+2S leq 3 Leftrightarrow boxedSleq 1$$

Consider the map $rcolonmathbbR^3longrightarrowmathbbR^3$ defined by $r(x,y,z)=(z,x,y)$. This is a rotation around the line $,xinmathbbR$, with angle $frac2pi3$. Therefore, if $(x,y,z)inmathbbR^3$, then the angle between $(x,y,z)$ and $r(x,y,z)$ is, at most, $frac2pi3$. So, if $a^2+b^2+c^2=1$ and if $theta$ is the angle between $(a,b,c)$ and $r(a,b,c)$,beginalignab+bc+ca&=bigllangle(a,b,c),r(a,b,c)bigrrangle\&=bigl|(a,b,c)bigr|^2costheta\&=costheta\&inleft[-frac12,1right].endalign