Prove that if $x \in \mathbb{R_{\ge 0}}$ Then $\left[\sqrt{x}\right]=\left[\sqrt{\left[x\right]}\right]$ Prove that if $x \in \mathbb{R_{\ge 0}}$ Then $$\left[\sqrt{x}\right]=\left[\sqrt{\left[x\right]}\right]$$
My Try:
Case $1.$ 
Let $\sqrt{x}=p$, where $p \in \mathbb{Z_{\ge 0}}$
Then $$LHS=p$$ and we have $x=p^2$ So
$$RHS=\left[\sqrt{\left[x\right]}\right]=\left[\sqrt{p^2}\right]=p=LHS$$
Case $2.$
Let $\sqrt{x}=p+f$ where $p \in \mathbb{Z_{\ge 0}}$ and $0 \lt f \lt 1$
We have $$x=p^2+2pf+f^2$$
$$LHS=\left[p+f\right]=p$$
We have $$RHS=\left[\sqrt{\left[p^2+f^2+2pf\right]}\right]=\left[\sqrt{p^2+\left[f^2+2pf\right]}\right]$$
Now we have $$f^2+2pf \gt 0$$
But how to prove $$f^2+2pf \lt 1$$
Note: I am not looking for a different solution. I need help to continue with my solution 
 A: $[\sqrt x]=m$ is equivalent $m\le \sqrt x<m+1$, i.e., to $m^2\le x<(m+1)^2$.
As $m^2$ is an integer, this implies $m^2\le [x]<(m+1)^2$, and hence $m\le \sqrt{[x]}<m+1$, i.e., $[\sqrt{[x]}]=m$.
A: The following specifically answers this part of OP's question: "I am not looking for a different solution. I need help to continue with my solution".

We have $$RHS=\left[\sqrt{\left[p^2+f^2+2pf\right]}\right]=\left[\sqrt{p^2+\left[f^2+2pf\right]}\right]$$
Now we have $$f^2+2pf \gt 0$$

Correct thus far.

But how to prove $$f^2+2pf \lt 1$$

That's not what you need to prove (and, in fact, doesn't hold true in general).
What has to be proved at this point is that:
$$
\begin{align}
\left\lfloor \sqrt{\lfloor x \rfloor} \right\rfloor = p \;\;&\iff\;\; \left\lfloor \sqrt{\left\lfloor (p+f)^2 \right\rfloor} \right\rfloor = p \\[5px]
 &\iff\;\; p \le \sqrt{\left\lfloor (p+f)^2 \right\rfloor} \lt p+1 \\[5px]
 &\iff\;\; \color{blue}{p^2 \le \left\lfloor (p+f)^2 \right\rfloor \lt (p+1)^2} \tag{1}
\end{align}
$$
But $0 \lt f \lt 1\,$, and therefore $p \lt p+f \lt p+1\,$, so:
$$
\begin{align}
p^2 \lt (p+f)^2 \lt (p+1)^2 \;\;&\implies\;\; \color{blue}{p^2} = \left\lfloor p^2 \right\rfloor \color{blue}{\le \left\lfloor (p+f)^2 \right\rfloor \lt} \left\lfloor (p+1)^2 \right\rfloor = \color{blue}{(p+1)^2} \tag{2}
\end{align}
$$
(The implication follows from the property of the greatest integer function that $\,a \lt b \implies \lfloor a \rfloor \le \lfloor b \rfloor\,$, with strict inequality $\,\lfloor a \rfloor \lt \lfloor b \rfloor\,$ if $\,b\,$ is an integer.)
The above proves $(2)\,$, which is identical with $(1)\,$, and therefore concludes the proof.
