Show $\frac{3997}{4001}>\frac{4996}{5001}$ I wish to show that $$\frac{3997}{4001}>\frac{4996}{5001}.$$
Of course, with a calculator, this is incredibly simple. But is there anyway of showing this through pure analysis? So far, I just rewrote the fractions:
$$\frac{4000-3}{4000+1}>\frac{5000-4}{5000+1}.$$

For all $n>0$, assume
$$\frac{4n-3}{4n+1}>\frac{5n-4}{5n+1}$$
$$(4n-3)(5n+1)>(4n+1)(5n-4)$$
$$20n^2-11n-3>20n^2-11n-4$$
$$-3>-4$$
which is true.
