Prove property of floor function (one with square roots) I want to prove that:
$$\lfloor\sqrt{x}\rfloor=\lfloor\sqrt{\lfloor x\rfloor}\rfloor$$
It's true that (by definition of floor operation):
$$\lfloor\sqrt{x}\rfloor\leq\sqrt{x}<\lfloor\sqrt{x}\rfloor+1$$
$$\lfloor\sqrt{\lfloor x\rfloor}\rfloor\leq\sqrt{\lfloor x\rfloor}<\lfloor\sqrt{\lfloor x\rfloor}\rfloor+1$$
But I don't know what comes next. Tried multiple conversions of those inequalities, but I did not see anything. Can you help? 

Let $x=i+f$ where $i\in \Bbb{Z},f\in[0,1)$ . 
$\therefore\sqrt x=\sqrt{i+f}$
Now $\exists g\in[0,1)$ such that $0<g<f$ and $f=2\sqrt ig+g^2$
Also $\exists h\in[0,1)$ such that $h>f$ and $f=-2\sqrt ih+h^2$
So $$\sqrt x=\sqrt i+g=\sqrt i-h$$
However $[\sqrt i]\leq [\sqrt i+g]\le [\sqrt i]+1$ and $[\sqrt i]-1\leq [\sqrt i-h]\le [\sqrt i]$
$$\therefore [\sqrt i+g]=[\sqrt i-h]=[\sqrt i]\\\implies [\sqrt x]=\left[\sqrt[2][x]\right]$$
