Proof: $ \lfloor \sqrt{ \lfloor x\rfloor } \rfloor = \lfloor\sqrt{x}\rfloor $. I need some help with the following proof:
$ \lfloor \sqrt{ \lfloor x\rfloor } \rfloor = \lfloor\sqrt{x}\rfloor $.
I got:
(1) $[ \sqrt{x} ] \le \sqrt{x} < [\sqrt{x}] + 1 $ (by definition?).
(2) $[ \sqrt{x} ]^2 \le x < ([\sqrt{x}] + 1)^2 $.
(3) $[ \sqrt{x} ]^2 \le [x] \le x < ([\sqrt{x}] + 1)^2$
??
 A: Note that every nonnegative number $x$ is between the squares of two consecutive nonnegative integers. Say $n^2 \leq x < (n+1)^2$. Then $n \leq \sqrt{x} < n + 1$, so $\lfloor \sqrt{x} \rfloor = n$ in this situation.
So you just have to make sure the left hand side is also $n$. Since $n^2$ is an integer, one has
$n^2 \leq \lfloor x \rfloor \leq x$, so $n^2 \leq \lfloor x \rfloor < (n+1)^2$. So exactly as above one has $\lfloor \sqrt{\lfloor x \rfloor} \rfloor = n$ as well.
A: Hint: for any positive integer $n$, $\sqrt{n+1}-\sqrt n=\frac 1{\sqrt{n+1}+\sqrt n}\lt 1$, hence 
$$\sqrt{x}\leqslant \sqrt{[x]+1}\lt 1+\sqrt{[x]},$$
hence $[\sqrt x]\leqslant \sqrt{[x]}$.
Since $[\sqrt x]$ is an integer, we actually have 
$$[\sqrt x]\leqslant [\sqrt{[x]}].$$
The opposite inequality is a consequence of increasingness of $t\mapsto \sqrt t$ and $[s]\leqslant s$ for any real number $s$.
A: First, you need that $x \geq 0$. Write $x=a+ \alpha$ where $\left\lfloor \sqrt x \right \rfloor=a$ and $\alpha <1$. Take $n^2 \leq a < (n+1)^{2}$ . Now, you need to prove that $n= \left \lfloor \sqrt a \right \rfloor = \left \lfloor \sqrt {a+ \alpha} \right \rfloor$. $\left \lfloor \sqrt {a+ \alpha} \right \rfloor \leq \left \lfloor \sqrt {(n+1)^2 -1 + \alpha} \right \rfloor <\left \lfloor \sqrt {(n+1)^2} \right \rfloor =n+1$. This implies $\left \lfloor \sqrt {a+ \alpha} \right \rfloor=n $ what you needed
