I'm currently reading a famous paper of Talagrand and fail to "easily see" that for a Gaussian process $(X_t)_{t\in T}$ and any $t_0 \in T$ one has
$$ E \sup_T | X_t | \leq E |X_{t_0}| + 2E \sup_T \, X_t $$
Could anybody please provide some details?
For reference: The statement appears in (1) on page 2 of Regularity of gaussian processes
First, as a small point of clarification: this holds for centered Gaussian processes, so $\mathbb{E}[X_t] = 0$ is constant.
Now, define $Y_t := X_t - X_{t_0}$. We have $$\mathbb{E}\left[\sup_{t \in T} |X_t|\right] \le \mathbb{E}[|X_{t_0}|] + \mathbb{E}\left[\sup_{t \in T} |Y_t|\right].$$ We write $Y = Y^+ - Y^-$, where $Y^+_t = Y_t 1_{Y_t \ge 0}$ and $Y^- = -Y_t 1_{Y_t \le 0}$. Because $Y$ is symmetric, $Y^+$ and $Y^-$ have the same law, so \begin{align*} \mathbb{E}\left[\sup_{t \in T} |Y_t|\right] &\le \mathbb{E}\left[\sup_{t \in T} Y^+_t\right] + \mathbb{E}\left[\sup_{t \in T} Y^-_t\right] \\ &= 2 \mathbb{E}\left[\sup_{t \in T} Y^+_t\right] \\ &= 2 \mathbb{E}\left[\sup_{t \in T} Y_t\right], \end{align*} where the last equality follows from the fact that $Y_{t_0} = 0$ so $\sup_{t \in T} Y_t = \sup_{t \in T} Y_t 1_{Y_t \ge 0}$. Finally, we have \begin{align*} \mathbb{E}\left[\sup_{t \in T} Y_t\right] &= \mathbb{E}\left[\sup_{t \in T} (X_t - X_{t_0})\right] \\ &= \mathbb{E}\left[\sup_{t \in T} X_t\right] - \mathbb{E}[X_{t_0}] \\ &= \mathbb{E}\left[\sup_{t \in T} X_t\right]. \end{align*} Combining everything, we thus conclude the desired inequality \begin{align*} \mathbb{E}\left[\sup_{t \in T} |X_t|\right] &\le \mathbb{E}[|X_{t_0}|] + \mathbb{E}\left[\sup_{t \in T} |Y_t|\right] \\ &\le \mathbb{E}[|X_{t_0}|] + 2 \mathbb{E}\left[\sup_{t \in T} Y_t \right] \\ &= \mathbb{E}[|X_{t_0}|] + 2 \mathbb{E}\left[\sup_{t \in T} X_t \right] \end{align*}