<s>[INST] How to prove that $ a^2+b^2+c^2=1$ implies $ab+bc+ca \in [-\frac12,1]$? 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.
 [/INST] 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:
$$1+2S = (a+b+c)^2 = (1\cdot a + 1\cdot b+1\cdot c)^2 \leq 3\cdot (a^2+b^2+c^2) = 3 $$$$\Leftrightarrow 1+2S \leq 3 \Leftrightarrow \boxed{S\leq 1}$$
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