Question

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

• A. $[\frac{1}{2},2]$
• B. $[0,2]$
• C. $[-\frac{1}{2},1]$
• D. $[-1,\frac{1}{2}]$

Answer 1

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

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

$(a+b+c{)}^2=a^2+b^2+c^2+2ab+2bc+2ca$

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

$\Rightarrow 1+2(ab+bc+ca)\ge 0\Rightarrow ab+bc+ca\ge -\frac{1}{2}$

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

$ab+bc+ca\le a^2+b^2+c^2\Rightarrow ab+bc+ca\le 1$

So answer is $[-\frac{1}{2},1]$

Hence, $\left(C\right)$