Question

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

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

Answer 1

The correct option is C

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

We have $a^2+b^2+c^2+2(bc+ca+ab)=(a+b+c{)}^2\ge 0$
$\Rightarrow 1+2(bc+ca+ab)\ge 0\Rightarrow bc+ca+ab\ge \frac{-1}{2}$

Also since A.M. $\ge$ G.M., for $a,b,c$we get
$\frac{b^2+c^2}{2}\ge \sqrt{b^2{c}^2}=bc,\frac{c^2+a^2}{2}\ge \sqrt{c^2{a}^2}=ca$ and
$\frac{a^2+b^2}{2}\ge \sqrt{a^2{b}^2}=ab$

Add all three inequalities to get
$\Rightarrow a^2+b^2+c^2\ge ab+bc+ca$
$\Rightarrow 1\ge ab+bc+ca$

So, the interval is $[\frac{-1}{2},1]$.