The below inequality is a more general case of Kolmogorov maximal inequality.
$\mathbf{Theorem}:$ Let $I \subset \mathbb{R}$ be an interval and $U, U_1, \dots, U_n$ be real RV's such that $U \in \mathbb{L}_1, \mathbb{P}[U \in I] =1$ and $\forall j=1, \dots, n \text{ holds that} \space U_j \le \mathbb{E}[U | U_1,\dots,U_j]$ a.s. and $\mathbb{P}[U \in I] =1$. Denote $M_n = \max[U_1,\dots,U_n]$. Then for $\psi:I\to[0,\infty]$ a nondecreasing convex function and an arbitrary $t \in I$ the following is true: $$\psi(t) \mathbb{P}[M_n > t] \le \int_{[M_n >t]} \psi(U) d \mathbb{P}$$
$\mathbf{Proof}:$ Let $t \in I$ and denote $\tau=\min\{i=1,2,\dots,n: U_i>t \}$. Clearly, $\tau$ is a stopping time w.r.t. the filtration $(\sigma(U_1,\dots,U_i),i \in \{1,\dots, n\}) $. and thus we obtain the estimate
\begin{align} \int_{[M_n >t]} \psi(U) d \mathbb{P} &=\sum_{i=1}^n \int_{[\tau=i]} \psi(U) d \mathbb{P}\\ &\ge \sum_{i=1}^n \int_{[\tau=i]} \psi(\mathbb{E}[U | U_1,\dots,U_i]) d \mathbb{P}\\ & \ge \sum_{i=1}^n \int_{[\tau=i]} \psi(U_i) d \mathbb{P}\\ & \ge \sum_{i=1}^n \mathbb{P}[\tau=i] \psi(t) \\ &=\psi(t) \mathbb{P}[M_n>t] \space \space \square \end{align}
$\mathbf{Q1}$ I don't understand how we come around to use the summation representation of the integral in the first line.
$\mathbf{A1}$ Since $U_i \le \mathbb{E}[U | U_1,\dots,U_i]$ and $[\tau=i]$ requires $U_i>t$ then we know that $t<\mathbb{E}[U | U_1,\dots,U_i]$. $M_n>t$ requires the $$\max_{i \in N, i \le n} U_i >t$$ So this way the stopping time is giving us the first (and the summation any subsequent) time that $M_n>t$. Smart.
$\mathbf{Q2}$ I don't get how we go from first to second. Jensen inequality? $$\mathbb{E}\psi(U) \ge \psi(\mathbb{E}U).$$ But I can't see any expectation in the first line.
Going from second to the third line merely exploits the fact, that $U_i \le \mathbb{E}[U | U_1,\dots,U_i]$ and the rest is easy.
I'd need a confirmation of my self-supplied A1 and an answer to Q2.
Q1: $M_n(\omega)>t$ if and only if there exists an $i\in \{1,\ldots,n\}$ such that $U_i(\omega)>t$. Hence $$ \{M_n>t\}=\bigcup_{i=1}^n \{U_i>t\}. $$ The problem here is that the union on the right-hand side is not a disjoint union, i.e. we can have that $U_i(\omega)>t$ and $U_j(\omega)>t$ for the same $\omega$. The way to go about this is to choose the minimum $i$ such that $U_i>t$, and hence $$ \{M_n>t\}=\bigcup_{i=1}^n \{\tau =i\}. $$ If you're still not convinced that the equality holds, you can just show the equality by showing both inclusions. Now that we have a disjoint union we can use that $$ \int_{\{M_n>t\}}\cdots\,\mathrm dP=\sum_{i=1}^n\int_{\{\tau=i\}}\cdots\,\mathrm dP. $$
Q2: Jensen's inequality for conditional expectations states that $\psi({\rm E}[U\mid \mathcal{F}_n])\leq {\rm E}[\psi(U)\mid \mathcal{F}_n]$ with $\mathcal{F}_n=\sigma(U_1,\ldots,U_n)$. Hence for $i=1,\ldots,n$ we have $$ \begin{align*} \int_{\{\tau=i\}}\psi({\rm E}[U\mid\mathcal{F}_n])\,\mathrm dP&\leq \int_{\{\tau=i\}}{\rm E}[\psi(U)\mid\mathcal{F}_n]\,\mathrm dP=\int_\Omega {\rm E}[\psi(U)\mid\mathcal{F}_n]1_{\{\tau=i\}}\,\mathrm dP\\ &=\int_\Omega {\rm E}[\psi(U)1_{\{\tau=i\}}\mid\mathcal{F}_n]\,\mathrm dP={\rm E}[{\rm E}[\psi(U)1_{\{\tau=i\}}\mid\mathcal{F}_n]]\\ &={\rm E}[\psi(U)1_{\{\tau=i\}}]=\int_{\{\tau=i\}}\psi(U)\,\mathrm dP. \end{align*} $$ Summing over $i=1,\ldots,n$ yields the desired result.