Proving a claim that $Z_n = \max\{(X_n^{(1)})^2, X_n^{(2)}\}$ is a submartingale Let $X_n^{(1)}, X_n^{(2)}, n \in \mathbb{N}$ be martingales. Let $\mathcal{F}_n$ be a filtration. I believe that $Z_n := \max\{X_n^{(1)}, X_n^{(2)}\}$ is a submartingale. I tried to show it by using Jensen's inequality since the mapping $(x,y) \mapsto \max\{x^2, y\}$ is convex. So,
\begin{align*}
\mathbb{E}[Z_n|\mathcal{F}_{n-1}] \geq \max\{\mathbb{E}[(X_n^{(1)})^2|\mathcal{F}_{n-1}], \mathbb{E}[X_n^{(2)}|\mathcal{F}_{n-1}]\} 
\end{align*}
Well, $\mathbb{E}[X_n^{(2)}|\mathcal{F}_{n-1}] = X_{n-1}^{(2)}$ by assumption. But I am not sure how to justify $\mathbb{E}[(X_n^{(1)})^2|\mathcal{F}_{n-1}] = (X_{n-1}^{(1)})^2$.

I think that just another application of Jensen's inequality does the trick:
\begin{align*}
\mathbb{E}[(X_n^{(1)})^2|\mathcal F_{n-1}] &\ge (\mathbb{E}[X_n^{(1)}|\mathcal F_{n-1}])^2 \\
&=(X_{n-1}^{(1)})^2,
\end{align*}
so
\begin{align*}
\mathbb{E}[Z_n|\mathcal F_{n-1}] &\ge \max\{\mathbb{E}[(X_n^{(1)})^2|\mathcal F_{n-1}],\mathbb{E}[X_n^{(2)}|\mathcal F_{n-1}]\} \\
&\ge \max\{(X_{n-1}^{(1)})^2, X_{n-1}^{(2)}\} \\
&= Z_{n-1}.
\end{align*}
