Sorry if I am being stupid. But I came to this lemma in the Appendix, written by Bourgain, Kontorovich and Magee, of the paper by Magee, Oh and Winter [MOW16]. It starts out like this:

Lemma 42 Let $\pi$ be a unitary $G$-representation on a Hilbert space $H$, and assume that the operator $A$ acts on $H$ via $$A\varphi=\sum_{j\in J}\pi(h_j)\varphi,$$ for some $h_j\in G$ and indexing set $J$. Assume that $A$ has the "spectral gap" property: there is some $C_0>0$ so that $$\langle A\varphi,\varphi\rangle\leq(1-C_0)|J|\|\varphi\|^2.$$ For sme positive coefficients $\kappa_j>0$, let $\widetilde {A}$ act on $H$ as $$\widetilde{A}\varphi=\sum_{j\in J}\kappa_j\pi(h_j)\varphi,$$ and assume that the $L^\infty$ norm of the coefficients is controlled by the $L^1$ norm, in the sense that for some $K\geq1$, $$\text{max}\kappa_j\leq K\bar{\kappa},$$ where $$\bar{\kappa}:=\frac{1}{|J|}\sum_{j}\kappa_j$$ is the coefficient average. Then $\widetilde{A}$ has the following "spectral gap": $$\langle\widetilde{A}\varphi,\varphi\rangle\leq\bar{\kappa}(1-C_0+\sqrt{K-1})|J|\|\varphi\|^2.$$

Here, the group $G$ is $\mathrm{SL}_2(\mathbb{Z}/q\mathbb{Z})$ with an arbitrary integer $q$.

In their paper, they said it was an exercise in Cauchy-Schwarz. However, I was wondering how they did it to make the final result consist of two terms, $(1-C_0)$ and $\sqrt{K-1}$, in the bracket.

I attempted a few times applying the Cauchy-Schwarz directly to the inner product, only to find terms with $K$ instead of $\sqrt{K-1}$.

This is a duplicate of the question asked in Mathstackexchange but without any further progress.

Any comments are appreciated.