Lower bound of Hecke eigenvalues of Maass form

If $x$ is large enough, then Rankin-Selberg theory will show that $S(x) \gg x^{1-\varepsilon}$. However, if $x$ is not large enough, then it is unknown how to obtain a lower bound for $S(x)$. In particular, it is unknown how to show that $S(x) \neq 0$. A good starting point for this is Chapter 13 of Iwaniec's book, Topics in Classical Automorphic Forms.

Following up on @Idoneal's comment, one can prove something a bit more precise. In particular, $S(x)\asymp\frac{x}{\log x}$, provided that $\log x\gg\log(\lambda N)$, where $f$ has level $N$ and Laplace eigenvalue $\lambda$. I'm assuming Matt Young's normalization and that the central character is unitary; also, I'm writing $f\asymp g$ to mean that both $f=O(g)$ and $g=O(f)$. This result follows from work of Motohashi https://arxiv.org/abs/1209.4140. A little extra effort is needed to get the claimed range, but all the necessary tools are in this paper.