$\lfloor \sqrt n+\sqrt {n+1}+\sqrt{n+2}+\sqrt{n+3}+\sqrt{n+4}\rfloor=\lfloor\sqrt {25n+49}\rfloor$ is true?

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 $n=0$).

This isn't a real proof, but note that: $$\sqrt{25n+49}=5\sqrt{n+2-1/25}\sim5\sqrt{n+2}$$ Which is quite close to the sum: $$\sum_{k+0}^4 \sqrt{n+k}$$ So as $n$ becomes larger the difference becomes smaller.

If you look at the series expansion around $n\to\infty$ of: $$f(n)=\sqrt{n}+\sqrt{n+1}+\sqrt{n+2}+\sqrt{n+3}+\sqrt{n+4}-\sqrt{25 n+49}$$ $$\sim\frac{1}{10n^{1/2}}+O\left(n^{-3/2}\right)$$ So it's clear that the only problem could arise if one side is equal to $M+\epsilon$, for some very small $\epsilon < 0.1n^{-1/2}$.

I would try and establish a lower bound for $\epsilon$ and try to arrive at a contradiction.

Let $$L(n):= \sqrt{n-2}+\sqrt{n-1}+\sqrt{n}+\sqrt{n+1}+\sqrt{n+2},\quad R(n):=\sqrt{25n-1}\ .$$ We have to show that $\lfloor L(n)\rfloor=\lfloor R(n)\rfloor$ for all $n\geq2$.

The function $$f(x):=\sqrt{1+x}+\sqrt{1-x}$$ has $f(0)=2$, $\>f'(0)=0$, and $$f''(x)=-{1\over4}\bigl((1+x)^{-3/2}+(1-x)^{-3/2}\bigr)\doteq-{1\over2}\quad\bigl(|x|\ll1\bigr)\ .$$ It follows that $$f(x)\doteq2-{x^2\over4}\quad\bigl(|x|\ll1\bigr)\ .$$ We therefore have $$L(n)=\sqrt{n}\left(1+f({1\over n})+f({2\over n})\right)\doteq\sqrt{n}\bigl(5-{5\over4n^2}\bigr)\doteq\sqrt{25n-{25\over 2n}}\ .\tag{1}$$ Assume now that $$n=k^2+\ell, \qquad 0\leq\ell<2k+1\ .$$ Then one can easily conclude from $(1)$ (with $R(n)$ its even simpler) that $$\lfloor L(n)\rfloor=\lfloor R(n)\rfloor=\cases{5k-1\quad&$(\ell=0)$\cr 5k&$(\ell\geq1)$\cr}\quad.$$