Question

If $a,b$ and $c$ are distinct positive real numbers and $a^2+b^2+c^2=1$, then $ab+bc+ca$ is-

• A. Less than $1$
• B. Equal to $1$
• C. Greater than $1$
• D. Any real number

Answer 1

Answer: (A)

Less than $1$

Given $a^2{+b}^2+c^2=1$
we know that, $(a+b+{c)}^2\ge 0$
$\Rightarrow a^2{+b}^2+c^2+2(ab+bc+ca)\ge 0$
$\Rightarrow 1+2(ab+bc+ca)\ge 0$
$\Rightarrow ab+bc+ca\ge \frac{-1}{2}$
Also, $(b-c{)}^2+(c-a{)}^2+(a-b{)}^2\ge 0$
$\Rightarrow {2(a}^2{+b}^2+c^2)-2(ab+bc+ca)\ge 0$
$\Rightarrow ab+bc+ca\le a^2{+b}^2+c^2$
$\Rightarrow ab+bc+ca\le 1$
Hence $\frac{-1}{2}\le ab+bc+ca\le 1$
Therefore, $ab+bc+ca$ is less than 1.