<s>[INST] 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.
 [/INST] 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: $3 \cdot \frac14 + \frac13 \cdot 4 > 3 \cdot \frac13 + \frac14 \cdot 4$.
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