We have $0<a<b$ which means $1/a>1/b$ and so $3+1/a>3+1/b$. Flipping over the fraction, we will get 1/(3 + 1/a) < 1/(3 + 1/b) and the inequality remains the same way round when 2 is added. Flipping over again for the last time we get

${\displaystyle \frac{1}{2+\frac{1}{3+\frac{1}{a}}}}$

is greater than

${\displaystyle \frac{1}{2+\frac{1}{3+\frac{1}{b}}}}$

The second part is a further expansion of the first, and in the process of repeating the above we know that it involves just one more flipping over of the fraction, thus

${\displaystyle \frac{1}{2+\frac{1}{3+\frac{1}{4+\frac{1}{a}}}}}$

is less than the same thing with $b$ in place of a as the inequality would be reversed again.

Lastly the continued fractions are expanded all the way down to $100+1/a$ and $100+1/b$. Observe the above process, we can tell that if the last or biggest number is odd then the continued fraction with a in it is bigger. If the last or biggest number is even then the continued fraction with $b$ in it is bigger. Each successive continued fraction involves one more 'flipping over' and reverses the inequality one more time. The following continued fraction is smaller than the same thing with $b$ in place of $a$:

${\displaystyle \frac{1}{2+\frac{1}{3+\frac{1}{4+\dots +\frac{1}{99+\frac{1}{100+\frac{1}{a}}}}}}}$