How to prove the $f(x) = \sqrt{x + \sqrt{x}}$ is injective. The function is 
$$
f(x)=\sqrt{x+\sqrt{x}}
$$
I know that you need to set up the equation $$\sqrt{x_1+\sqrt{x_1}}=\sqrt{x_2+\sqrt{x_2}}$$
and you have to solve step by step until you get $x_1=x_2$. But I am having difficulty figuring out what to do.  I have tried squaring both sides, factoring, then completing the square but nothing has worked. All answers appreciated.  Please keep in mind that I need every step, thank you.
 A: Can you use the fact that the (nonnegative branch of the) square root is strictly increasing?
If so, one way to prove it would be to notice that if for some $0\leq x_1< x_2$ we have 
$\sqrt{x_1+\sqrt{x_1}}=\sqrt{x_2+\sqrt{x_2}}$
then
$x_1+\sqrt{x_1}=x_2+\sqrt{x_2}$
and so
$x_1-x_2=\sqrt{x_2}-\sqrt{x_1}$
but since we are assuming $x_1<x_2$, the left side of the equation is negative, and the right side is positive.

Just in case, to prove that the square root is strictly increasing, let $x_1,x_2$ be nonnegative real numbers with $x_1<x_2$. If we suppose that $\sqrt{x_1}\geq \sqrt{x_2}$ then, since we are dealing with nonnegative numbers, squaring the inequality gives us $x_1\geq x_2$ which contradicts our hypothesis.
A: The calculus way: We can compute $f'$ easily enough to find that if $f(x) = \left(x + x^{1/2}\right)^{1/2}$ then
$$f'(x) = \frac{1}{2} \left(x + x^{1/2}\right)^{-1/2} \left(1 + \frac{1}{2} x^{-1/2}\right)$$
This is positive for all $x > 0$, so the function is one-to-one.
A: Hint
If $x_1 <x_2$ then $\sqrt{x_1} < \sqrt{x_2}$. Adding the two inequalities you get
$x_1+\sqrt{x_1} < x_2+\sqrt{x_2}$. Finally take the square root:
$$\sqrt{x_1+\sqrt{x_1}} < \sqrt{x_2+\sqrt{x_2}} \,.$$
