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
$$\sqrt**{\prod_{i=1}^{b} \frac{c+i}{a+i}} \le \frac{1}{b} \sum_{i=1}^{b} \frac{c+i}{a+i} \quad \text{ i.e. } \quad \prod_{i=1}^{b} \frac{c+i}{a+i} \le \left( \frac{1}{b} \sum_{i=1}^{b} \frac{c+i}{a+i} \right)^{b}.$$
This establishes the inequality on the right. Swapping the roles of $a$ and $c$, then taking the reciprocal, gives the left hand inequality.