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?

This is actually a very devious inequality problem disguised as a calculus problem, and it took me very long to realise this. But eventually what gave it away was that calculating the second derivative of $f(x)$ would be a monstrous task, and that the $\sqrt{\dfrac{ax}{ax+8}}$ term seemed very suspicious (As we shall see, dividing both numerator and denominator by $ax$ would yield something very interesting and useful.)
First let us rename $x$ as $b, b \geq 0$. The case where $b=0$ is trivial, since we have $\dfrac{1}{\sqrt{1+a}} + \dfrac{1}{\sqrt{1+b}} + \sqrt{\dfrac{ab}{ab+8}} = \dfrac{1}{\sqrt{1+a}} +1 $. Thus, we may assume $a,b >0$. Let $a=2x,b=2y$, where $x, y >0$.
Thus, $$\dfrac{1}{\sqrt{1+a}} + \dfrac{1}{\sqrt{1+b}} + \sqrt{\dfrac{ab}{ab+8}} =\dfrac{1}{\sqrt{1+2x}} + \dfrac{1}{\sqrt{1+2y}} + \dfrac{1}{\sqrt{1+\dfrac{2}{xy}}}.$$
Let $2z=\dfrac{2}{xy} \Rightarrow xyz=1$. So really, this problem is equivalent to us proving the following inequality:
Inequality:
Let $x,y,z \in \mathbb{R^+}$, $xyz=1$. Prove that $1<\dfrac{1}{\sqrt{1+2x}} + \dfrac{1}{\sqrt{1+2y}} + \dfrac{1}{\sqrt{1+2z}} <2. $
Now, the lower bound is relatively easy, but the upper bound is a real killer.
Lower Bound:
Since $1+2x >1 , \sqrt{1+2x} < 1+2x \Rightarrow \dfrac{1}{\sqrt{1+2x}} > \dfrac{1}{1+2x}$.  Thus it suffices for us to prove that:
\begin{align}
\sum_{\text{cyc}} \dfrac{1}{1+2x} \geq 1 & \iff \dfrac{(1+2y)(1+2z) + (1+2x)(1+2z) + (1+2x)(1+2y)}{(1+2x)(1+2y)(1+2z)} \geq 1 \\
& \iff 3+4(x+y+z) + 4(yz+xz+xy) \geq 9+2(x+y+z) +  4(yz+xz+xy)  \\
& \iff 2(x+y+z) \geq 6 \\
& \iff x+y+z \geq 3 \\
\end{align}
Which follows immediately from AM-GM and the given condition.
Upper Bound:
The following proof borrowed an important idea introduced in Yufei Zhao's inequalities notes: Inequalities. Sometimes switching the roles of the constraint and the inequality can yield wonders!
We begin by letting $p=\dfrac{1}{\sqrt{1+2x}}, q=\dfrac{1}{\sqrt{1+2y}}, r=\dfrac{1}{\sqrt{1+2z}} $, and $p,q,r<1$. We have to prove that $xyz=1 \Rightarrow p+q+r <2$. Instead of proving this directly, we will prove the contrapositive: $p+q+r \geq 2 \Rightarrow xyz <1 \Rightarrow xyz \neq 1$. Next we carry out the following algebraic manipulations:
\begin{align}
p=\dfrac{1}{\sqrt{1+2x}} & \Rightarrow p^2=\dfrac{1}{1+2x} \\
& \Rightarrow \dfrac{1}{p^2} = 1+2x \\
& \Rightarrow x=\dfrac{1-p^2}{2p^2} = \dfrac{(1+p)(1-p)}{2p^2} < \dfrac{1-p}{p^2}\\
\end{align}
But
\begin{align}
p+q+r \geq 2 & \Rightarrow p \geq 2-q-r \\
& \Rightarrow 1-p \leq -1+q+r = qr-(1-q)(1-r) < qr \\
& \Rightarrow x < \dfrac{qr}{p^2}.
\end{align}
Similarly, we obtain that $y < \dfrac{pr}{q^2}$ and $z < \dfrac{pq}{r^2}$. Thus, $xyz <\dfrac{qr}{p^2} \cdot \dfrac{pr}{q^2} \cdot  \dfrac{pq}{r^2} =1$, and we are done.
Final note: In fact, I believe that we have the following stronger upper bound:
$$\dfrac{1}{\sqrt{1+2x}} + \dfrac{1}{\sqrt{1+2y}} + \dfrac{1}{\sqrt{1+2z}} \leq  \sqrt{3}.$$
However, I have absolutely no clue on how to proceed. Perhaps someone else can come up with a nice idea?
