Solve $\lfloor \sqrt x +\sqrt{x+1}+\sqrt{x+2}\rfloor=x$ $$\lfloor \sqrt x +\sqrt{x+1}+\sqrt{x+2}\rfloor=x$$ 
I tried to solve this equation.
   First thing is $\lfloor \sqrt x +\sqrt{x+1}+\sqrt{x+2}\rfloor \in \mathbb{Z} $ so $x \in \mathbb{Z}$
second $$\sqrt x +\sqrt{x+1}+\sqrt{x+2} \geq \sqrt 0 +\sqrt{0+1}+\sqrt{0+2} \\\to x \in \mathbb{N}$$ so we can check $x=1,2,3,4,5,6,7,8,9,\ldots$ by a MATLAB program.  I checked the natural numbers to find solution. I found $x=8,9$ worked here.
Now my question is about somehow an analytical solving of the equation, or another idea. Can any one help me? Thanks in advance.  
 A: This is a little plodding, but $x\in\mathbb{Z}$ and $x\ge0$ (required in order for $\sqrt x$ to be real) tells us
$$x\ge\lfloor\sqrt0+\sqrt1+\sqrt2\rfloor=\lfloor2.414\rfloor=2$$
which tells us
$$x\ge\lfloor\sqrt2+\sqrt3+\sqrt4\rfloor=\lfloor5.145\rfloor=5$$
which tells us
$$x\ge\lfloor\sqrt5+\sqrt6+\sqrt7\rfloor=\lfloor7.331\rfloor=7$$
which tells us
$$x\ge\lfloor\sqrt7+\sqrt8+\sqrt9\rfloor=\lfloor8.474\rfloor=8$$
Finally, we see that if $x\ge10$, then the function $f(x)=\sqrt x+\sqrt{x+1}+\sqrt{x+2}-x$ is negative, since $f(10)=-0.056997...$ and
$$f'(x)={1\over2\sqrt x}+{1\over2\sqrt{x+1}}+{1\over2\sqrt{x+2}}-1\lt{3\over2\sqrt 9}-1=-{1\over2}\lt0$$
so we cannot have $x\le\sqrt x+\sqrt{x+1}+\sqrt{x+2}$, which is required in order to have $x=\lfloor\sqrt x+\sqrt{x+1}+\sqrt{x+2}\rfloor$, for $x\ge10$.  This leaves the two possibilities $x=8$ and $9$, which do solve the equation.
A: You can use inequalities to simplify your problem. 
Since $\lfloor x \rfloor \le x$. Therefore we've 
\begin{align}
x&= \lfloor \sqrt x+\sqrt {x+1}+\sqrt{x+2} \rfloor \\
&\le \sqrt x+\sqrt {x+1}+\sqrt{x+2} \\
&\le 3\sqrt {x+2}\\
\end{align}
$$\implies x^2 \le 9(x+2) \; ; x \in \mathbb Z$$
This gives us the range $x \in [-1,10] \tag1$.
Also $\lfloor x \rfloor \ge x-1$. Therefore we've
\begin{align}
x&= \lfloor \sqrt x+\sqrt {x+1}+\sqrt{x+2} \rfloor \\
&\ge  \sqrt x+\sqrt {x+1}+\sqrt{x+2} \color{red}{-1}\\
&\ge 3\sqrt {x}-1\\
\end{align}
$$\implies x+1 \ge 3\sqrt x$$
$$\implies x^2-7x+1\ge 0 \, ; x \in \mathbb Z$$
This gives us $x \in (-\infty, 0]\cup [7,\infty)\tag2$ 
Taking intersection of $(1)$ and $(2)$, and taking care of domain, I.e. $x\ge 0$, we get
$$\color{blue}{x \in \{0,7,8,9,10\}}$$
Now you can check for $x=0,7,8,9,10$, which is quite easy now.
A: First solve the real inequalities
$$
x \le \sqrt{x}+\sqrt{x+1}+\sqrt{x+2} < x+1
\tag{1}$$
Solution of $x = \sqrt{x}+\sqrt{x+1}+\sqrt{x+2}$ is numerically $9.8956$ and solution of $x = \sqrt{x}+\sqrt{x+1}+\sqrt{x+2}=x+1$ is numerically $7.9813$.  So solution of (1) is
$$
7.9813 < x \le 9.8956
$$
Finally, assume $x$ is an integer.  We get $x=8$ or $x=9$.
A: Side comment:  $\frac{\sqrt{m-1} + \sqrt{m+1}}2 \ge \sqrt{\sqrt{m^2 - 1}}>\sqrt m$ by $AM-GM$ theorem.
So $\sqrt{x} + \sqrt{x+1} + \sqrt{x+2} \approx 3\sqrt{x+1}$ but slightly larger.
$x \le \sqrt{x} + \sqrt{x+1} + \sqrt{x+2}< 3\sqrt{x+1} < x+1$ implies $x^2 < 9x + 9 < x^2 + 2x +1$ (the first inequality is definite, the second is aproximate).
$x^2 <9x+9$ (which is definitely true) implies $x^2 - 9x -9 = (x -\frac {9+\sqrt{81 + 4*9}}2)(x -\frac {9-\sqrt{81 + 4*9}}2) < 0$ which implies $x < \frac {9+\sqrt{81 + 4*9}}2= \frac {9+3\sqrt{13}}2\approx 9.9$ but is *definitely less than $10$.  $9$ is a definite upper limit of $x$
$9x + 9 < x^2 + 2x + 1$ (which is only approximate) implies $x^2 - 7x -8 = (x-8)(x+1) > 0$ implies $x > 8$ (or $x < 1$ which would not be possible.)  
So $9$ is a solution and $8$ is possible if $\sqrt{x} + \sqrt{x+1} + \sqrt{x+2} < 9 \le 3\sqrt{x+1}$ ... which, ... actually is the case for $x= 8$. (As $3\sqrt{8+1} = 9$ exactly.)
So solutions are $8$ and $9$.
Hmmm.... I haven't actually shown it fails for $x < 8$ but for that to hold the difference between $\sqrt{x} + \sqrt{x+1} + \sqrt{x+2}$ and $3\sqrt{x+1}$ must be larger than $1$ and that's .... well, clearly not true for $x \approx 8$.  We can just check that $7$ fails.
I'm also assuming the rate of which $x$ increases will "outstrip" $\sqrt{x} + \sqrt{x+1} + \sqrt{x+2}$.  Calculus verifies this. 
A: It is clear that for some $x$ you have $LHS\lt x$ and for other $x$ you have $LHS\gt x$ and besides that you could  have the equality for small enough $x$.
A straightforward calculation gives you  $LHS\lt x$ for $x\lt8$ and $LHS\gt x$ for  $\gt9$. 
The solutions are $x=8$ and $x=9$ because $\sqrt8+\sqrt9+\sqrt{10}=8.9907...$ and $\sqrt9+\sqrt{10}+\sqrt{11}=9.4789...$ 
A: Here's a way
to get bounds on $x$
for a generalized version of this.
We want to solve
$x
=\lfloor \sum_{k=1}^n \sqrt{x+a_k} \rfloor
$.
Upper bound:
If
$a = \max(a_k)$,
then
$\begin{array}\\
x
&=\lfloor \sum_{k=1}^n \sqrt{x+a_k} \rfloor\\
&\le\lfloor \sum_{k=1}^n \sqrt{x+a} \rfloor\\
&=\lfloor n \sqrt{x+a} \rfloor\\
&\le n \sqrt{x+a} \\
\text{so}\\
x^2
&\le n^2(x+a)\\
\end{array}
$
Therefore
$x^2-n^2x+n^4/4
\le n^2(a+n^2/4)
$
or
$|x-n^2/2|
\le n\sqrt{a+n^2/4}
= (n/2)\sqrt{4a+n^2}
$
so
$x
\le (n/2)(n+\sqrt{4a+n^2})
$.
For this case,
$n=3, a=2$,
so
$x
\le (3/2)(3+\sqrt{17})
\lt 10.7
$
so
$x \le 10$.
Lower bound:
If
$a = \min(a_k)$,
then
$\begin{array}\\
x
&=\lfloor \sum_{k=1}^n \sqrt{x+a_k} \rfloor\\
&\ge\lfloor \sum_{k=1}^n \sqrt{x+a} \rfloor\\
&=\lfloor n \sqrt{x+a} \rfloor\\
&\gt n \sqrt{x+a}-1 \\
\text{so}\\
(x+1)^2
&\gt n^2(x+a)\\
&= n^2(x+1+a-1)\\
\end{array}
$
Therefore
$(x+1)^2-n^2(x+1)+n^4/4
\gt n^2(a-1+n^2/4)
$
or
$|x+1-n^2/2|
\gt n\sqrt{a-1+n^2/4}
= (n/2)\sqrt{4a-4+n^2}
$
so
$x+1-n^2/2
\gt (n/2)\sqrt{4a-4+n^2}
$
or
$x+1-n^2/2
\lt -(n/2)\sqrt{4a-4+n^2}
$.
Therefore
$x
\gt n^2/2-1+(n/2)\sqrt{4a-4+n^2}
$
or
$x
\lt n^2/2-1-(n/2)\sqrt{4a-4+n^2}
$.
For this case,
$n=3, a=0$,
so
$(n/2)\sqrt{4a-4+n^2}
=\frac32\sqrt{-4+9}
=\frac32\sqrt{5}
\approx 3.35
$
and
$n^2/2-1
=3.5
$
so
$x > 6.85$
or
$x < 0$.
Therefore
$x \ge 7$,
so that,
combining the bounds,
$7 \le x \le 10$.
A: Just narrowed the possible solutions using inequality of integer at the both side
$\left[\sqrt{{x}}+\sqrt{{x}+\mathrm{1}}+\sqrt{{x}+\mathrm{2}}\right]={x} \\ $
$\mathrm{3}\sqrt{{x}}\leqslant{x}\leqslant\mathrm{3}\sqrt{{x}+\mathrm{2}} \\ $
$\mathrm{9}{x}\leqslant{x}^{\mathrm{2}} \leqslant\mathrm{9}{x}+\mathrm{18} \\ $
${x}\in{N}\Rightarrow{x}\in\left\{\mathrm{9},\mathrm{10}\right\} \\ $
