Expectation over a max operation Let $X \in \mathbb{R}_{\geq 0}$ be a "non-negative" random variable and $c$ is a "given" strictly positive number. I wonder if the following inequality holds:
$$
E[\max\{X,c\}] \leq \max\{E[X],c\},
$$
where $E[\cdot]$ is the expectation.
I suspect from Jensen's inequality the other way around should be true; but since the $c$ above is a certain constant, I'm still (naively) hopeful that it could be true.
 A: Similar to winperikle's answer, just tightening the arguments a bit:
$\max\{X, c\} \geq X$ and $\max\{X, c\} \geq c$. So, by taking expectation, $\text{E}\left(\max\{X, c\}\right) \geq \text{E} X$ and $\text{E}\left(\max\{X, c\}\right) \geq c$. Combining, we get $\text{E}\left(\max\{X, c\}\right) \geq \max \{\text{E} X, c\}$. 
These arguments can be generalized to show that for a sequence of $\mathcal{L}_1$ random variables $(X_n)_{n\geq 1}$, $\text{E} \left(\sup_{n \geq 1} |X_n| \right) \geq \sup_{n \geq 1} \text{E}|X_n|$.  
A: The inequality you have asserted is false: A simple counter-example is $X \sim \text{Bin}(2,\tfrac{1}{2})$ and $c=1$, which gives you the expectation:
$$\mathbb{E}(\max(X,c)) = \frac{3}{4} \cdot 1 + \frac{1}{4} \cdot 2 = \frac{5}{4}.$$
For this counter-example we have:
$$\frac{5}{4} = \mathbb{E}(\max(X,c)) > \max(\mathbb{E}(X),c) = 1.$$

There is a related inequality that is true: Although the inequality you have asserted is false (or at least, not generally true), the following alternative inequality is true:
$$\mathbb{E}(\max(X,c)) \geqslant \max(\mathbb{E}(X), c).$$
This inequality can easily be proven either for the discrete or continuous (or mixed) case.  For a discrete random variable you have:
$$\begin{equation} \begin{aligned}
\mathbb{E}(\max(X,c)) 
&= \sum_{x \in \mathscr{X}} \max(x,c) \cdot p_X(x) \\[8pt]
&\geqslant \sum_{x \in \mathscr{X}} x \cdot p_X(x) = \mathbb{E}(X). \\[8pt]
\end{aligned} \end{equation}$$
You also have:
$$\begin{equation} \begin{aligned}
\mathbb{E}(\max(X,c)) 
&= \sum_{x \in \mathscr{X}} \max(x,c) \cdot p_X(x) \\[8pt]
&\geqslant \sum_{x \in \mathscr{X}} c \cdot p_X(x) = c. \\[8pt]
\end{aligned} \end{equation}$$
Putting these together gives the inequality.
A: Let X be uniform in (0, 5) and c=2. Here you have a counterexample with each side of the inequality being 3.5 and 2.5
A: If $\text{max}(\mathbb{E}[X], c) = c$, as $\text{max}(X,c) \geq c$, we have 
\begin{align*}
\mathbb{E}[\text{max}(X,c)] &\geq c \\
&\geq \text{max}(\mathbb{E}[X],c)
\end{align*}
When $\text{max}(\mathbb{E}[X],c) = \mathbb{E}[X]$ then again as $\text{max}(X,c) \geq X$ we have
\begin{align*}
\mathbb{E}[\text{max}(X,c)] &\geq \mathbb{E}[X] \\
&\geq \text{max}(\mathbb{E}[X],c)
\end{align*}
So that the inequality is actually the other way
$$
\mathbb{E}[\text{max}(X,c)] \geq \text{max}(\mathbb{E}[X], c)
$$
