Question

If $a^2+b^2+c^2=1$, then the range of $ab+bc+ca$ is:

• A. $[-\frac{1}{2},1]$
• B. $[-\frac{1}{2},\infty )$
• C. $[1,\infty )$
• D. None

Answer 1

Answer: (A)

$[-\frac{1}{2},1]$

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

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

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

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

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

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

Again, ${(a+b+c)}^2\ge 0$ ..... ($\because$ It is a squared term)

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

$\Rightarrow 1+2(ab+bc+ca)\ge 0$ ..... (substituting $a^2+b^2+c^2=1$)

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

$\therefore -\frac{1}{2}\le ab+bc+ca)\le 1$

$\therefore$ Range $=[-\frac{1}{2},1]$