Models of cosmological inflation are plagued with a severe and seemingly unavoidable problem: in order to produce density perturbations of an amplitude consistent with large-scale observations, the self-coupling $\lambda$ of the inflaton field has to be tuned to an excessively small value. In all these models, however, the scalar field is taken to be minimally coupled to the scalar curvature ($\xi =0\mathrm{}$). It is shown here that in the more general case of nonminimal coupling ($\xi \ne 0$), and within the framework of Linde's chaotic inflation, the constraint on the self-coupling could be relaxed by several orders of magnitude. We are led to this conclusion by the combination of two key results. (1) Contrary to previous common belief, the curvature coupling $\xi$ can be almost arbitrarily large without upsetting the inflationary scenario. In fact, the larger $\xi$ is, the better the model behaves. (2) Considerations regarding the amplitude of density perturbations constrain the ratio $\frac{\lambda }{{\xi }^2}$ rather than $\lambda$. Thus, by a suitable choice of $\xi$, the self-coupling $\lambda$ can be made as large as desired. It is found that for large $\xi$ the amplitude of density perturbations is much smaller than in $\xi =0\mathrm{}$ models: ${\left(\delta \frac{\rho }{\rho }\right)|}_{\xi >1\mathrm{}}\approx {\left(48N{\xi }^2\right)}^{-\frac{1}{2}}{\left(\frac{\delta \rho }{\rho }\right)|}_{\xi =0\mathrm{}}$, where $N\sim 70$. For example, this represents a drop of over 4 orders of magnitude for $\xi ={10}^3$. This same value results in a dramatic 9 orders of magnitude weakening of the constraint on $\lambda$ according to our formula ${\lambda }_{\mathrm{constraint}|\xi >1\mathrm{}}\approx 48N{\xi }^2{\lambda }_{\mathrm{constraint}|\xi =0\mathrm{}}$. Non-minimal coupling may thus provide a relatively simple solution to the long-standing problem of excessive density perturbations in inflationary models.

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