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.

If $a<b$ then
$$(a+x)(b-x)$$
is increasing in x for $0\leq x \leq \frac{b-a}{2}$.
Using this $(2+1/2)*(5+1/5)$ is larger than $(2+1/5)*(5+1/2)$ and $(3+1/3)*(4+1/4)$ is larger than $(3+1/4)*(4+1/3)$.
Intuitively, the square maximises the area over all rectangles with the same circumference. To maximise a product where the factors have a fixed sum, we must try to get the factors as close as possible.
