IRL_ABooklet

The fraction in the middle can also be written this way $$x=\frac{a!(b+c)!}{(a+b)!c!}=\frac{(c+1)(c+2)\cdots (c+b)}{(a+1)(a+2)\cdots (a+b)}=\prod _{i=1}^b\frac{c+i}{a+i}.$$ Because $1+\frac{c-a}{a+i}=\frac{c+i}{a+i}$, we have $$1+\frac{c-a}{b}\sum _{i=1}^b\frac{1}{a+i}=\frac{1}{b}\sum _{i=1}^b\left(1+\frac{c-a}{a+i}\right)=\frac{1}{b}\sum _{i=1}^b\frac{c+i}{a+i}.$$ The AM-GM inequality gives $$\sqrt[b]{\prod _{i=1}^b\frac{c+i}{a+i}}\le \frac{1}{b}\sum _{i=1}^b\frac{c+i}{a+i}\text{ i.e. }\prod _{i=1}^b\frac{c+i}{a+i}\le {\left(\frac{1}{b}\sum _{i=1}^b\frac{c+i}{a+i}\right)}^b.$$ This establishes the inequality on the right. Swapping the roles of $a$ and $c$, then taking the reciprocal, gives the left hand inequality.