Question

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

• A. $[1,\frac{2}{3}]$
• B. $[-\frac{1}{2},1]$
• C. $[-1,\frac{1}{2}]$
• D. $[2,-4]$

Answer 1

The correct option is B $[-\frac{1}{2},1]$

$(a+b+c{)}^2\ge 0\therefore (a+b+c{)}^2=a^2+b^2+c^2+2(ab+bc+ca)\ge 0$ . . . (i)
$\therefore$ 1 + 2 (ab + bc + ca) $\ge$ 0 $(a^2+b^2+c^2=1)$
$\Rightarrow (ab+bc+ca)\ge -\frac{1}{2}$ . . . (ii)
Also, $a^2+b^2\ge 2ab,b^2+c^2\ge 2bc,c^2+a^2\ge 2ac(a^2+b^2+b^2+c^2+c^2+a^2)\ge 2ab+2bc+2ac\Rightarrow a^2+b^2+c^2\ge ab+bc+ac(\because a^2+b^2+c^2=1)\Rightarrow ab+bc+ac\le 1\therefore -\frac{1}{2}\le ab+bc+ac\le 1.$