Question

If $a^2+b^2+c^2=1,$ then $ab+bc+ca$ lies in the interval

• A. $\left[\frac{1}{2},2\right]$
• B. $[-1,2]$
• C. $\left[-\frac{1}{2},1\right]$
• D. $\left[-1,\frac{1}{2}\right]$

Answer 1

Answer: (C)

Given, $a^2+b^2+c^2=1$ ...(i)
${(a+b+c)}^2\ge 0$
$\Rightarrow$ $a^2+b^2+c^2+2(ab+bc+ca)\ge 0$
$\Rightarrow$ $1+2(ab+bc+ca)\ge 0$
$\Rightarrow$ $ab+bc+ca\ge -\frac{1}{2}$ ....(ii) and
${(a-b)}^2\ge 0$
$\Rightarrow$ $a^2+b^2-2ab\ge 0$
$\Rightarrow$ $a^2+b^2\ge 2ab$ Similarly, $b^2+c^2\ge 2bc,c^2+a^2\ge 2ca$
$\therefore$ $2(a^2+b^2+c^2)\ge 2ab+2bc+2ca$
$\Rightarrow$ $a^2+b^2+c^2\ge ab+bc+ca$
$\Rightarrow$ $1\ge ab+bc+ca$ ...(iii)
Hence, from Eqs. (ii) and (iii),
$-\frac{1}{2}\le ab+bc+ca\le 1$
$\therefore$ Interval $=\left[-\frac{1}{2},1\right]$