<s>[INST] 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.
 [/INST] 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$$x+1<\sqrt{x^2+3x+3}<x+2.$$
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