For the given function $f(x)=\frac{1}{\sqrt {1+x}} +\frac{1}{\sqrt {1+a}} + \sqrt{\frac{ax}{ax+8}}$, prove that $1
For the given function $f(x)=\frac{1}{\sqrt {1+x}} +\frac{1}{\sqrt {1+a}} + \sqrt{\frac{ax}{ax+8}}$, prove that $1<f(x)<2$ for +ve $a$ and $x\ge 0$
I tried the most rudimentary method ie. differentiating wrt x and hoped to get at least 2 solutions for $f’(x)$, unfortunately I wasn’t able to find a solution for implying function is monotonous . Now whether it’s sis increasing or decreasing still remains a mystery, given that we don’t know what $a$ is.
What method can be used to solve this?

I think that the derivative gives a good indication.
$$f(x)=\frac{1}{\sqrt {1+x}} +\frac{1}{\sqrt {1+a}} + \sqrt{\frac{ax}{ax+8}}$$
$$f'(x)=4 \sqrt{\frac{a}{x (a x+8)^3}}-\frac{1}{2 (x+1)^{3/2}}$$ Without any constraint, the derivative cancels at four points
$$x_{1,2}=\pm \frac{2 \sqrt{2}}{\sqrt{a}}\qquad x_{3,4}=\frac{2 \left(16-6a\pm \sqrt{2(a-8)^2 (2-a)}\right)}{a^2}$$ $x_{3,4}$ do not exist in the real domain if $a >2$ but, in any case, remains $x_{1}= \frac{2 \sqrt{2}}{\sqrt{a}}$.
On the other side $f''(x_1)$ is positive if $0 <a <2$ and negative if $a>2$.
Add all of that to @mathcounterexamples's answer.
