The domain of the function $\sqrt{\frac{1}{\sqrt{x}}+\sqrt{x+1}}$ is :

The given function $f\left(x\right)=\sqrt{\frac{1}{\sqrt{x}}+\sqrt{x+1}}$ is defined for $x>0$        …(i)

For domain 

$\begin{array}{r}\frac{1}{\sqrt{x}}+\sqrt{x+1}\ge 0\\ \Rightarrow \sqrt{x^2+x}\ge -1\\ \Rightarrow x^2+x-1\ge 0\end{array}$

Solving this as quadratic equation, we get, 

$\begin{array}{l}x=\frac{-1\pm \sqrt{5}}{2}\\ \Rightarrow \left[x-\left(\frac{-1-\sqrt{5}}{2}\right)\right]\left[x-\left(\frac{-1+\sqrt{5}}{2}\right)\right]\ge 0\\ \Rightarrow \left[x-\left(\frac{-1-\sqrt{5}}{2}\right)\ge 0\text{ or }\left[x-\left(\frac{-1+\sqrt{5}}{2}\right)\right]\ge 0\right.\\ \Rightarrow x\ge \frac{-1-\sqrt{5}}{2}\text{ or }x\ge \frac{-1+\sqrt{5}}{2}\end{array}$

and $\left[x-\left(\frac{-1-\sqrt{5}}{2}\right)\right]\le 0\text{ or }\left[x-\left(\frac{-1+\sqrt{5}}{2}\right)\right]\le 0$

$\Rightarrow x\le \frac{-1-\sqrt{5}}{2}\text{ or }x\le \frac{\sqrt{5}-1}{2}\dots \left(3\right)$

from all these condition we get 

$0\le x\le \frac{\sqrt{5}-1}{2}$