Let
$$
S=\sum_{T<t\leq2T}\sum_{R<r\leq2R}b_tc_r e(\alpha r^2t^2).
$$
Applying the <Cauchy-Schwarz inequality> in $t$ gives
$$
|S|^2
\leq\|b\|_2^2
\sum_t\left|\sum_r c_re(\alpha r^2t^2)\right|^2
=\|b\|_2^2
\sum_{r_1,r_2}c_{r_1}\overline{c_{r_2}}
\sum_t e\bigl(\alpha t^2(r_1^2-r_2^2)\bigr).
$$
Take absolute values and apply Cauchy-Schwarz to $(r_1,r_2)$. Since $\sum_{r_1,r_2}|c_{r_1}c_{r_2}|^2=\|c\|_2^4$, expanding the remaining square gives
$$
|S|^4
\leq\|b\|_2^4\|c\|_2^4
\sum_{\substack{T<t_1,t_2\leq2T\\R<r_1,r_2\leq2R}}
e\bigl(\alpha(t_1^2-t_2^2)(r_1^2-r_2^2)\bigr).
$$
Taking fourth roots proves the claimed <bilinear quadratic exponential sum fourth-moment bound>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 117 / 3 / c / Solution

Two applications of the <Cauchy-Schwarz inequality> give
$$
\left|\sum_{t,r}b_tc_re(\alpha r^2t^2)\right|
\leq\|b\|_2\|c\|_2
\left(\sum_{t_1,t_2,r_1,r_2}
e\bigl(\alpha(t_1^2-t_2^2)(r_1^2-r_2^2)\bigr)
\right)^{1/4}.
$$
The fourth moment exposes differences of squares that can be factored and estimated by one-dimensional geometric sums.

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 Incoming links: Bilinear quadratic exponential sum fourth-moment bound