Question

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

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

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

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

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

Also $a^2+b^2+c^2-(ab+bc+ca)$

= $\frac{1}{2}\left[\right(a-b{)}^2+(b-c{)}^2+(c-a{)}^2]\ge 0$

$\Rightarrow$ 1 - ( ab + bc +ca ) $\ge$ 0

$\Rightarrow$ ab +bc + ca $\ge$ 0 .....(2)

Combining (1) and (2) , we ger

$-\frac{1}{2}\le ab+bc+ca\le 1$ .

i.e., ab + bc + ca lies in $[\frac{-1}{2},1]$