<s>[INST] If $\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}=\frac{3}{2}\sqrt{\frac{x}{x+\sqrt{x}}}$, how can it imply $x\ge1$? I encountered this problem in a book. Solve the equation $\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}=\frac{3}{2}\sqrt{\frac{x}{x+\sqrt{x}}}$.
According to the book, $x\ge1$ from observation. And after some algebra, they get $\sqrt{x}(2\sqrt{x}-1-2\sqrt{x-1})=0$. But since $\sqrt{x}\ge1>0$, so $2\sqrt{x}-1-2\sqrt{x-1}=0$ and therefore $x=\frac{25}{16}$.
What I didn't get is why is $x\ge1$? From what we can tell is $x+\sqrt{x}>0$ and $x-\sqrt{x}\ge0$ but from these two how can we conclude $x\ge1$? 
Any help is appreciated, thanks!
 [/INST] By itself, the expression on the left is also defined at $x=0$, but then the expression on the right is not. 
If $0\lt x\lt 1$, then $\sqrt{x}\gt x$, so $x-\sqrt{x}$ is negative. Presumably we are working in the reals, so there is no such thing as $\sqrt{x-\sqrt{x}}$. 
</s>