= Solution
Taking the <Fourier transform> of the equation in the sense of <distributions> gives
$$
(|\xi|^4-|\xi|^2+1)\widehat u(\xi)=\widehat f(\xi).
$$
The <Fourier multiplier>
$$
m(\xi)=|\xi|^4-|\xi|^2+1=\left(|\xi|^2-\frac12\right)^2+\frac34
$$
is everywhere positive. Define $\widehat u=\widehat f/m$. Since
$$
\sup_{\xi\in\mathbb R^n}\frac{(1+|\xi|^2)^2}{m(\xi)}<\infty,
$$
we have
$$
\lVert u\rVert_{H^{s+4}}^2
=\int(1+|\xi|^2)^{s+4}\frac{|\widehat f(\xi)|^2}{m(\xi)^2}\,d\xi
\leq C^2\lVert f\rVert_{H^s}^2.
$$
Thus $u\in H^{s+4}$ exists and obeys the required estimate. If two solutions existed, their difference would have $m\widehat u=0$; positivity of $m$ proves uniqueness.
A <classical solution> needs continuous derivatives through order four. Applying the <Sobolev embedding theorem> with $k=4$ to $u\in H^{s+4}$ shows that this is guaranteed when
$$
s+4>4+\frac n2,
$$
or equivalently
$$
\boxed{s>\frac n2}.
$$
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