This is from a Gaokao question I found online but can't find a solution for:
$$ f(x) = \frac{1}{\sqrt{1+x}} + \frac{1}{\sqrt{1+a}} + \sqrt{\frac{ax}{ax+8}} $$
$$ \text{Prove that for positive reals} \, a,x \, \text{that} \, 1 \le f(x) \le 2 $$
For me the intuition seems simple enough - for low values of x or a, the first two terms approach 1 while the third term approaches 0. Meanwhile for high values of a and x, the third term approaches 1 while the first two terms shrink, but still retain sufficient value to keep f(x) above 1 as their asymptotic approach to 0 is slower than the third term's approach to 1.
What approaches can prove this rigorously?
Link to the original Gaokao post: https://www.quora.com/Are-there-any-examples-of-hard-Chinese-Gaokao-math-questions