Show that $f(x)$ always has value between 1 and 2

Question

Given $f(x) = \frac{1}{\sqrt{1 + x}} + \frac{1}{\sqrt{1 + a}} + \sqrt{\frac{ax}{ax + 8}}$ for $x \in (0, +\infty)$. How do I prove that $\forall a \geq 0$, $1 < f(x) < 2$?