Proving $\left\lfloor \sqrt{x^2+3x+3}\right\rfloor = x+1$ 
Let $$f(x)=\sqrt{x^2+3x+3}$$
Claim: $$\lfloor f(x)\rfloor = x+1$$

I came across this expression in a question and made this claim based on observations
$$⌊f(1)⌋=\lfloor\sqrt{7}\rfloor=2$$
$$⌊f(2)⌋=\lfloor\sqrt{13}\rfloor=3$$
$$⌊f(3)⌋=\lfloor\sqrt{21}\rfloor=4$$
$$⌊f(4)⌋=\lfloor\sqrt{31}\rfloor=5$$
and so on ...
However, I can't think of a formal proof for my claim.
It would be appreciated if someone could help me with the proof or give some kind of hint.
Thank you.
 A: I am assuming that $x\in\Bbb Z_+$.
Note that $x^2+3x+3=x^2+2x+1+x+2=(x+1)^2+x+2$. So, $x^2+3x+3>(x+1)^2$. Could it be that $x^2+3x+3\geqslant(x+2)^2$? No, because\begin{align}x^2+3x+3\geqslant(x+2)^2&\iff x^2+3x+3\geqslant x^2+4x+4\\&\iff3x+3\geqslant4x+4.\end{align}So$$(x+1)^2<x^2+3x+3<(x+2)^2$$and therefore$$x+1<\sqrt{x^2+3x+3}<x+2.$$
A: The statement $[\sqrt{x^2 +3x + 3}] = x+1$ is true if and only if

*

*$x+1$ is an integer and

*$x+1 \le\sqrt{ x^2 + 3x + 3}^2 < x+2$.

The first is true if and only if $x$ is an integer.  SO this not true if $x \not \in \mathbb Z$.  But if $x \in\mathbb Z$ this might be true.
As for 2)
$(x+1)^2 = x^2 +2x + 1$ and $(x+2)^2 = x^2 + 4x + 4$.
At first blush it looks like we could say
As $x^2 + 2x + 1 < x^2 + 3x + 3 < x^2 + 4x + 4$ we always have $x+1 < \sqrt{x^2 + 3x + 3} < x+2$....
But there are two mistakes with that:
i)  $x^2 + 2x + 1 < x^2 + 3x + 3 < x^2 + 4x + 4\iff x+2 > 0$ and $x+1 > 0$.  If $x < -1$ this assumption isn't true and
ii)  $M^2 < x^2 < N^2 \implies M < x < N$ only if all the terms are non-negative.
So if $x \in \mathbb Z$ and if $x > -1$ (or in other words if $x \in \mathbb N \cup \{0\}$ then that is a valid proof that $[\sqrt{x^2 +3x + 3}] = x+1$.
But if $x \not \in Z$ or if $x \le -1$ then that statement is not true.
.....
P.S.
If $x \le -2$ with $x\in \mathbb Z$ then we have
$(x+2)^2 = x^2 + 4x + 4 < x^2 +3x + 3 < x^2 + 2x + 1=(x+1)^2$ so
$-x - 2 = |x+2|< \sqrt{x^2 + 3x + 3} < |x+1| = -x-1$ so
$[ \sqrt{x^2 + 3x + 3}] = -x -2$.
If $x = -1$ then $[ \sqrt{x^2 + 3x + 3}]=[\sqrt{1-3+3}]=1= x+ 2=-x$
P.P.S
If $x\in \mathbb R$ and $x > -1$ we still have $x + 1 < \sqrt{x^2 + 3x + 3} < x+2$ so if $x \not \in \mathbb Z$ we have $[x]+1< x+1 < \sqrt{x^2 + 3x + 3} < x+ 2 < \lceil x\rceil + 2 \le [x] + 3$.  So $[ \sqrt{x^2 + 3x + 3}] = \begin{cases} [x]+1\\ [x] + 2\end{cases}$ depending upon whether there is an integer between $[x]+1$ and $x + 1$ or not.  (i.e. if $x \ge [x] +\frac 12$ or if $x < [x]+\frac 12$.
We can do a similar calculation for if $x < -2; x\not \in \mathbb Z$ but I don't see any great incentive to pin down that much detail.
