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 \[ ab+bc+ca < 1. \] 

Answer 2

An alternate approach uses the AM-GM style inequality \(xy\le \dfrac{x^2+y^2}{2}\) applied to each pair of the three numbers, checked against every option.

1. Option A (less than 1): By the inequality \(xy\le\frac{x^2+y^2}{2}\) applied to each pair, \[ ab\le\frac{a^2+b^2}{2},\quad bc\le\frac{b^2+c^2}{2},\quad ca\le\frac{c^2+a^2}{2}. \] Adding these, \[ ab+bc+ca\le\frac{2a^2+2b^2+2c^2}{2}=a^2+b^2+c^2=1. \] Equality would require \(a=b=c\) in every pairwise inequality, but \(a,b,c\) are given to be distinct, so the inequality must be strict: \(ab+bc+ca<1\). This supports option A.
2. Option B (equal to 1): Equality is only possible when \(a=b=c\), which is excluded since the numbers are distinct, so this option is ruled out.
3. Option C (greater than 1): This directly contradicts the inequality \(ab+bc+ca\le1\) proved above, so it is false.
4. Option D (any real number): Since \(a,b,c\) are positive, \(ab+bc+ca>0\) always, and the inequality above further restricts it below \(1\); it cannot be an arbitrary real number.

The pairwise inequality strictly bounds \(ab+bc+ca\) below \(1\) whenever \(a,b,c\) are distinct, so option A is the only one consistent with this.

Hence, the correct answer is option A: less than \(1\).