Question

If $a^2+b^2+c^2=1$, find the range of $ab+bc+ca$.

Answer 1

Given, $a^2+b^2+c^2=1$

We know that, $(a+b+c{)}^2\ge 0$

$\Rightarrow a^2+b^2+c^2+2(ab+bc+ca)\ge 0$

$\Rightarrow 1+2(ab+bc+ca)\ge 0$

$\Rightarrow ab+bc+ca\ge \frac{-1}{2}$

Again, $(b-c{)}^2+(c-a{)}^2+(a-b{)}^2\ge 0$

$\Rightarrow 2(a^2+b^2+c^2)-2(ab+bc+ca)\ge 0$

$\Rightarrow ab+bc+ca\le a^2+b^2+c^2$

$\Rightarrow ab+bc+ca\le 1$

Hence, $\frac{-1}{2}\le ab+bc+ca\le 1$