Comparing $(2+\frac{1}{2})(3+\frac{1}{3})(4+\frac{1}{4})(5+\frac{1}{5})$ with $(2+\frac{1}{5})(3+\frac{1}{4})(4+\frac{1}{3})(5+\frac{1}{2})$ This question appeared in one of the national exams (MCQs) in Saudi Arabia.
In this exam;

*

*Using calculators is not allowed,

*The student have $72$ seconds on average to answer one question.


PROBLEM:
Compare $a=(2+\frac{1}{2})(3+\frac{1}{3})(4+\frac{1}{4})(5+\frac{1}{5})$ with $b=(2+\frac{1}{5})(3+\frac{1}{4})(4+\frac{1}{3})(5+\frac{1}{2})$.
CHOICES:
A) $a>b$
B) $a<b$
C) $a=b$
D) Given information is not enough

Using algebra to evaluate each expression is easy, and the correct choice is $A$, but that will take a long time.
Any suggestion to solve this problem in a short time? THANKS.

The sum of all four factors is the same in both cases. To maximize the product, we want the factors to be as close together as possible.
The factors $(2 + \frac12)(5 + \frac15)$ are closer to their average than $(2 + \frac15)(5 + \frac12)$, so $(2+\frac12)(5+\frac15) > (2+\frac15)(5+\frac12)$.
Similarly, $(3 + \frac13)(4+\frac14) > (3 + \frac14)(4 + \frac13)$.

We could also compare each of these pairs by multiplying them out. But we don't have to multiply out everything. When we expand $(2 + \frac12)(5 + \frac15)$ and $(2 + \frac15)(5 + \frac12)$, the terms $2\cdot 5$ and $\frac12 \cdot \frac15$ will be common between them. However, $2 \cdot \frac15 + \frac12 \cdot 5 > 2 \cdot \frac12 + \frac15 \cdot 5$, which is not hard to see: just $\frac12 \cdot 5$ is $2.5$ on the left, and the right is $2$.
The same thing happens with the other pair: $3 \cdot \frac14 + \frac13 \cdot 4 > 3 \cdot \frac13 + \frac14 \cdot 4$.
