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

For any two reals, \(xy\le\dfrac{x^2+y^2}{2}\). Applying this to each of the three pairs and adding:

1. \(ab\le\dfrac{a^2+b^2}{2}\)
2. \(bc\le\dfrac{b^2+c^2}{2}\)
3. \(ca\le\dfrac{c^2+a^2}{2}\)

Adding gives \(ab+bc+ca\le a^2+b^2+c^2=1\), with equality only if \(a=b=c\). Since \(a,b,c\) are distinct, the inequality is strict, so \(ab+bc+ca\) is less than \(1\).