How to prove that $ a^2+b^2+c^2=1$ implies $ab+bc+ca \in [-\frac12,1]$?

Question

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.

Answer 1

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

For the lower bound: $$(a+b+c)^2 \geq 0 \Leftrightarrow a^2+b^2+c^2 + 2S = 1 + 2S \geq 0 \Leftrightarrow \boxed{S \geq -\frac{1}{2}}$$ For the upper bound we use Cauchy-Schwarz: $$1+2S = (a+b+c)^2 = (1\cdot a + 1\cdot b+1\cdot c)^2 \leq 3\cdot (a^2+b^2+c^2) = 3 $$$$\Leftrightarrow 1+2S \leq 3 \Leftrightarrow \boxed{S\leq 1}$$

Answer 2

Consider the map $r\colon\mathbb{R}^3\longrightarrow\mathbb{R}^3$ defined by $r(x,y,z)=(z,x,y)$. This is a rotation around the line $\bigl\{(x,x,x)\,|\,x\in\mathbb{R}\bigr\}$, with angle $\frac{2\pi}3$. Therefore, if $(x,y,z)\in\mathbb{R}^3$, then the angle between $(x,y,z)$ and $r(x,y,z)$ is, at most, $\frac{2\pi}3$. So, if $a^2+b^2+c^2=1$ and if $\theta$ is the angle between $(a,b,c)$ and $r(a,b,c)$,\begin{align}ab+bc+ca&=\bigl\langle(a,b,c),r(a,b,c)\bigr\rangle\\&=\bigl\|(a,b,c)\bigr\|^2\cos\theta\\&=\cos\theta\\&\in\left[-\frac12,1\right].\end{align}