Question

If $a,b,c$ 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

Hint:

When comparing $ab+bc+ca$ with $a^2+b^2+c^2$, use $(a-b)^2+(b-c)^2+(c-a)^2\ge0$ to get a sharp bound.

Answer 1

The Correct Option is A

Use the identity \[ (a-b)^2+(b-c)^2+(c-a)^2 =2\,(a^2+b^2+c^2)-2\,(ab+bc+ca)\ \ge 0. \] Hence \(ab+bc+ca \le a^2+b^2+c^2=1\). 
Equality holds only when \(a=b=c\), which is impossible here (they are distinct). Therefore \[ ab+bc+ca < 1. \]