$\lfloor \sqrt n+\sqrt {n+1}+\sqrt{n+2}+\sqrt{n+3}+\sqrt{n+4}\rfloor=\lfloor\sqrt {25n+49}\rfloor$ 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 $\lfloor \sqrt n+\sqrt {n+1}+\sqrt{n+2}+\sqrt{n+3}+\sqrt{n+4}\rfloor=\lfloor\sqrt {25n+49}\rfloor$
 A: 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.
A: 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
$$\lfloor L(n)\rfloor=\lfloor R(n)\rfloor=\cases{5k-1\quad&$(\ell=0)$\cr 5k&$(\ell\geq1)$\cr}\quad.$$
A: 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 $n=0$).
A: I want to proof this only by using basic arithmetic and therefore avoiding Jensen's Inequality and Taylor's Theorem.
It's simpler if one replaces $n$ by $n-2$. On then gets
$$ \lfloor \sqrt{n-2}+\sqrt{n-1}+\sqrt n+\sqrt {n+1}+\sqrt{n+2}\rfloor=\lfloor\sqrt {25n-1}\rfloor$$
It is necessary to show that
$$ \sqrt {25n-1} \lt \sqrt{n-2}+\sqrt{n-1}+\sqrt n+\sqrt {n+1}+\sqrt{n+2} \tag{1} $$
$$ \sqrt{n-2}+\sqrt{n-1}+\sqrt n+\sqrt {n+1}+\sqrt{n+2}  \lt \sqrt {25n} \tag{2}$$
We first prove
$$ \sqrt{x-b}+\sqrt{x+b} \lt \sqrt{x-a}+\sqrt{x+a} , 0 \lt a \lt b \lt x \tag{3}$$
by squaring $(3)$ we get
$$ \sqrt{x-b}\sqrt{x+b} \lt \sqrt{x-a}\sqrt{x+a} $$
and by squaring this again we get 
$$x^2-b^2 \lt x^2-a^2 $$
and therefore 
$$ a^2 \lt b^2 $$
The arguments is also valid in the opposite direction , because all numbers we squared where positive.
$$ \sqrt{n-2}+\sqrt{n-1}+\sqrt n+\sqrt {n+1}+\sqrt{n+2} \lt 5 \sqrt{n}$$ 
follows immediately from 
$$\sqrt{n-2}+\sqrt{n+2}=\sqrt{n}+\sqrt{n}$$
which can be deduced from $(3)$ and $b=2$ and $a=0$, 
and from 
$$\sqrt{n-1}+\sqrt{n+1}=\sqrt{n}+\sqrt{n}$$
which can be deduced from $(3)$ and $b=1$ and $a=0$, 
So $(2)$ is proofen.
To proof $(1)$ we notice that
$$ \sqrt{n-2}+\sqrt{n-1}+\sqrt n+\sqrt {n+1}+\sqrt{n+2} \gt 2\sqrt{n-2}+\sqrt n+ 2 \sqrt{n+2} $$
because
$$\sqrt{n-1}+ \sqrt{n+1} \gt \sqrt{n-2}+ \sqrt{n+2}$$
So it is sufficiont to check if 
$$ 2 \sqrt{n-2}+\sqrt n+ 2 \sqrt{n+2} \gt  \sqrt{25n-1} $$
We square
$$ 2 \sqrt{n-2} + 2 \sqrt{n+2} \gt  \sqrt{25n-1} - \sqrt{n} $$
and get
$$8\,\sqrt{n-2}\,\sqrt{n+2}+8\,n>-2\,\sqrt{n}\,\sqrt{25\,n-1}+26\,n-1$$
and bring the squareroot terms to the LHS and the non non-squareroot terms  to the RHS of the inequation
$$ 2\,\sqrt{n}\,\sqrt{25\,n-1}+8\,\sqrt{n-2}\,\sqrt{n+2} \gt 18\,
 n-1 $$
We repeate this process until we get
$$9216\,n^3-185664\,n^2+20544\,n-66049 \gt 0$$
The polynomial can be written as 
$$\left(n-21\right)\,\left(9216\,n^2+7872\,n+185856\right)+3836927$$
so for $n \ge 21$ the polynomial is positive and $(1)$ is valid.
