Let $g_1, ..., g_n \stackrel{\text{iid}}{\sim} N(0,1)$ and $T$ be an arbitrary index set for a bounded family of distributions, say bounded by $M$.
Let $Y_i(t), i \in \{1, ..., n\}, t \in T$ be iid from the family and also independent of the Gaussians.
I want to show the following: $$E \left(\sup_{t \in T} \left \lvert \sum_{i=1}^n g_i Y_i(t) \right \rvert \right) \leq 2E\left(\sup_{t \in T} \sum_{i=1}^n g_i Y_i(t) \right) + C\sqrt{n}$$
where $C$ is a fixed constant.
I really don't know where to start, or which symmetrization principle I'm supposed to use (if any).
I know that for $\epsilon_i$ iid Bernoulli's and independent of everything else that the LHS of the inequality is equal to $$E\left(\sup_{t \in T} \left \lvert \sum_{i=1}^n \epsilon_i |g_i| Y_i(t) \right \rvert \right)$$ but I don't know any tricks on how to leverage this.
Since asking, I produced another rather feeble attempt (perhaps it's useful to someone). Denote $g \cdot Y(t)$ the corresponding dot product of vectors and decompose the absolute value into pieces:
$$\begin{aligned} LHS & = E\left(\sup_{t \in T} \bigg( g \cdot Y(t) + 2(-g \cdot Y(t) \mathbb{1}(g \cdot Y < 0) \bigg) \right)\\ & \leq E\left(\sup_{t \in T} \sum_{i=1}^n g_i Y_i(t) \right) + 2 E\left(\sup_{t \in T} - g \cdot Y(t) \mathbf{1}(g \cdot Y(t) < 0) \right)\end{aligned}$$