Let us define the following sequence:
$$ x_{n + 1} = -4 - \frac{6}{x_n},$$
with first term $x_1 = -4$. This sequence naturally arises in numerical methods of approximations. With simple C++ program I have realised some interesting property:
$$ \frac{1}{n^2} \leq |x_n| \leq n^2,\mbox{ for any } 2 \leq n \leq 10^8. $$
But I am still unable to proof these inequalities or even establish similar ones. May be you know some good ways to work with such a sequences?
This is not an answer, but I would like to give a possible track by showing that a linear algebra approach is fruitful.
Let $x_n=\dfrac{N_n}{D_n}$ (numerators and denominators). The given relationship can thus be written under the form:
$$\dfrac{N_{n+1}}{D_{n+1}}=\dfrac{-4N_n-6D_n}{N_n}$$
i.e. under a matrix form:
$$\begin{pmatrix}N_{n+1}\\D_{n+1}\end{pmatrix}=\underbrace{\begin{pmatrix}-4&-6\\ \ \ 1& \ \ 0\end{pmatrix}}_{M}\begin{pmatrix}N_n\\D_n\end{pmatrix}$$
The existence of a (real valued !) fixed point is clearly equivalent to the existence of a (real valued) eigenvector for $M$. But this is not possible because the two eigenvalues of matrix $M$ are complex : $-2\pm \sqrt{2}i$.
Diagonalizing matrix $M$ and taking it at the power $n$ gives the general formula that one finds in the answer of @Claude Leibovici.
With $x_n=y_{n+1}/y_n$, we get the linear recurrence
$$y_{n+2} = -4y_{n+1} - 6y_n,$$ giving the explicit solution
$$x_n=\frac{(-2+\sqrt{-2})^{n+1}-(-2-\sqrt{-2})^{n+1}}{(-2+\sqrt{-2})^n-(-2-\sqrt{-2})^n}=\sqrt{6}\,\frac{\sin(n+1)\phi}{\sin n\phi},$$
where $\cos\phi=-2/\sqrt{6}$ and $\sin\phi=1/\sqrt{3}$. That means $\cos2\phi=1/3$, so $\phi$ can't be a rational multiple of $\pi$: if we had $\phi=p\pi/q$ with integer $p,q$, that would mean $$1=\cos2p\pi=\cos2q\phi=T_q(\cos2\phi)$$ with the $q$th Chevbyshev polynomial $T_q$, i.e. the polynomial equation $T_q(x)-1=0$ should have the solution $x=1/3$. That's possible only if $3$ divides the leading coefficient, but that's $2^{q-1}$.
So it's clear $\sin n\phi$ can't be zero for $n\neq0$, but it can come arbitrarily close, since the fractional parts of $n\phi/\pi$ are dense in $[0,1]$. So the inequality of the OP would mean that the irrationality measure of $\phi/\pi$ is $\le3$, it looks ambitious to prove that.
This is not an answer.
After Professor Vector's comment, I worked (using a CAS) the recurrence relation (not the most pleasant one !).
The final result is $$x_n= \frac{2 i \sqrt{2} \left(2-i \sqrt{2}\right)^n}{\left(2-i \sqrt{2}\right)^n-\left(2+i \sqrt{2}\right)^n}-(2+i \sqrt{2})$$ which simplifies to $$x_n=-2-\sqrt{2} \cot \left(n \cot ^{-1}\left(\sqrt{2}\right)\right)$$ and there is no integer solution for $$n \cot ^{-1}\left(\sqrt{2}\right)=k \pi$$
Concerning the bounds, I am stuck.
