$\lfloor \sqrt n+\sqrt {n+1}+\sqrt{n+2}+\sqrt{n+3}+\sqrt{n+4}\rfloor=\lfloor\sqrt {25n+49}\rfloor$ is true? I found the following relational expression by using computer:
For any natural number $n$, 
$$\lfloor \sqrt n+\sqrt {n+1}+\sqrt{n+2}+\sqrt{n+3}+\sqrt{n+4}\rfloor=\lfloor\sqrt {25n+49}\rfloor.$$
Note that $\lfloor x\rfloor$ is the largest integer not greater than $x$.
I can neither prove this nor find any counterexample even by using computer. 
Could you show me how to prove this? Or could you get an counterexample?
Update: I've just asked a related question.
Generalization of $\lfloor \sqrt n+\sqrt {n+1}+\sqrt{n+2}+\sqrt{n+3}+\sqrt{n+4}\rfloor=\lfloor\sqrt {25n+49}\rfloor$

Note that by the strict concavity of $\sqrt{x}$, Jensen's Inequality says
$$
\hspace{-1cm}\sqrt{n}+\sqrt{n+1}+\sqrt{n+2}+\sqrt{n+3}+\sqrt{n+4}\lt5\sqrt{n+2}=\sqrt{25n+50}\tag{1}
$$
More precisely, Taylor's Formula with remainder says
$$
\begin{align}
\sqrt{n+2+x}
=\sqrt{n+2}+\frac{x}{2\sqrt{n+2}}-\int_0^x\frac{(x-t)\,\mathrm{d}t}{4\sqrt{n+2+t}^3}\tag{2}
\end{align}
$$
Summing $(2)$ for $x\in\{-2,-1,0,1,2\}$ yields
$$
\hspace{-1cm}5\sqrt{n+2}-\frac5{4n^{3/2}}\lt\sqrt{n}+\sqrt{n+1}+\sqrt{n+2}+\sqrt{n+3}+\sqrt{n+4}\lt5\sqrt{n+2}\tag{3}
$$
Since
$$
\begin{align}
\sqrt{25n+50}-\sqrt{25n+49}
&=\frac1{\sqrt{25n+50}+\sqrt{25n+49}}\\
&\gt\frac1{2\sqrt{25n+50}}\\
&\gt\frac5{4n^{3/2}}\quad\text{for }n\ge14\tag{4}
\end{align}
$$
we get that for $n\ge14$,
$$
\hspace{-5mm}\sqrt{25n+49}\lt\sqrt{n}+\sqrt{n+1}+\sqrt{n+2}+\sqrt{n+3}+\sqrt{n+4}\lt\sqrt{25n+50}\tag{5}
$$
$(5)$ says that for $n\ge14$,
$$
\left\lfloor\sqrt{n}+\sqrt{n+1}+\sqrt{n+2}+\sqrt{n+3}+\sqrt{n+4}\right\rfloor
=\left\lfloor\sqrt{25n+49}\right\rfloor\tag{6}
$$
It is simple to verify $(6)$ for $1\le n\lt14$ (it is false for $n=0$).
