Show $\frac{3997}{4001}>\frac{4996}{5001}$

Subtract $1$ from both sides:

$$-\frac{4}{4001} > -\frac{5}{5001}$$

Remove the negatives and flip the inequality:

$$\frac{4}{4001} < \frac{5}{5001}$$

This is equivalent to

$$\frac{1}{1000.25} < \frac{1}{1000.2}$$

Multiply both sides by $1000.25 \cdot 1000.2$ to get

$$1000.2 < 1000.25$$

Since this is true, the original inequality must be true.

Continuing with your approach, write

$\frac{4000(1-3/4000)}{4000(1+1/4000)}>\frac{5000(1-4/5000)}{5000(1+1/5000)}$

Then cancel on both sides and cross multiply:

$(1-\frac{3}{4000})(1+\frac{1}{5000})>(1+\frac{1}{4000})(1-\frac{4}{5000})$

Now, expand both sides:

$1+\frac{1}{5000}-\frac{3}{4000}-\frac{3}{20000000}>1+\frac{1}{4000}-\frac{4}{5000}-\frac{4}{20000000}$

$\frac{5}{5000}-\frac{4}{4000}+\frac{1}{20000000}>0$

$\frac{1}{20000000}>0$.

Since the above is clearly true, then the original statement is true also.

First a minor but cute detour.

$a_n=\frac {n}{n*1000 +1} = \frac 1{1000 + 1/n} $.

As $n $ gets larger $1/n $ gets smaller, so $a_n $ get larger. So $a_1 < a_2 < a_3 <.... $.

So $\frac {3997}{4001} = 1 - \frac 4 {4001} = 1 - a_4 > 1 - a_5 = 1 - \frac 5 {5001}=\frac {4996}{5001} $.

Okay, this wasn't as straightforward or as easy as the others but in my opinion it was i) informative as why it ought to be true and ii) cute.