<s>[INST] $\lfloor \sqrt n+\sqrt {n+1}+\sqrt{n+2}+\sqrt{n+3}+\sqrt{n+4}\rfloor=\lfloor\sqrt {25n+49}\rfloor$ 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 $\lfloor \sqrt n+\sqrt {n+1}+\sqrt{n+2}+\sqrt{n+3}+\sqrt{n+4}\rfloor=\lfloor\sqrt {25n+49}\rfloor$
 [/INST] 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.
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