Question

If the product of three positive real numbers is 1 and their sum is greater than sum of their reciprocals, then

• A. exactly one of them exceeds 1
• B. each of them equals 1
• C. exactly one of them is less than 1
• D. None of these

Answer 1

The correct option is A exactly one of them exceeds 1

Let the numbers be a, b, $\frac{1}{ab}$. We are given
$a+b+\frac{1}{ab}>\frac{1}{a}+\frac{1}{b}+abNow,(a-1)(b-1)(\frac{1}{ab}-1)=1+(a+b+\frac{1}{ab})-(ab+\frac{1}{a}+\frac{1}{b})-1>0$
$\Rightarrow$ either all three of a-1, b-1, $\frac{1}{ab}-1$ are positive or exactly one of them is positive.
But all three of a-1, b-1, $\frac{1}{ab}-1$ cannot be positive.