<s>[INST] Finding range of function $$f(x) = \frac{1}{\sqrt{1+x}}+\frac{1}{\sqrt{1+a}}+\sqrt{\frac{ax}{ax+8}}$$
Prove that for all positive real number $a$, $1<f(x)<2$
According  to me i think question is not correct.
as at $a= 16$, we have case when function reaches
infinite value in left of $-1/2$.
 [/INST] You are right, it seems to be true for
$$f(x) = \frac{1}{\sqrt{1+x}}+\frac{1}{\sqrt{1+a}}+\sqrt{\frac{ax}{ax+8}}$$
only for $x>0$, indeed
$$f'(x) = -\frac12\frac{1}{\sqrt[2]{(1+x)^3}}+\frac{a}{ax+8}\frac{1-\frac{ax}{ax+8}}{2\sqrt{\frac{ax}{ax+8}}}<0$$
and $$f(0)=1+\frac{1}{\sqrt{1+a}}<2, \quad f(x)\to 1 \quad x\to \infty$$
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