Question

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

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

Answer 1

Answer: (D)

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

given $a^2+b^2+c^2=1$
$(a+b+c{)}^2=a^2+b^2+c^2+2(ab+bc+ca)$
$=1+2(ab+bc+ca)$
$\Rightarrow \frac{(a+b+c{)}^2-1}{2}=ab+bc+ca$
minimum value of $(a+b+c{)}^2$ is zero.
$\Rightarrow minvalueofab+bc+ca=\frac{-1}{2}$...(1)
Now consider the expression
$(a-b{)}^2+(b-c{)}^2+(c-a{)}^2$
The above expression is $\ge 0$
$\Rightarrow 2(a^2+b^2+c^2-ab-bc-ca)\ge 0$
$\Rightarrow ab+bc+ca\le a^2+b^2+c^2$
$\Rightarrow ab+bc+ca\le 1$....(2)
From (1) and (2), the range of $ab+bc+ca$ is
$[\frac{-1}{2},1]$