$\mathbf{E}[\max\{X, a\}] \geq \max\{\mathbf{E}[X], a\}$ 
Let $X$ be a random variable with finite mean, (i.e., $\mathbf{E}[X] < \infty$), and $a \in \mathbb{R}$. Prove that
$$ \mathbf{E}[\max\{X, a\}] \geq \max\{\mathbf{E}[X], a\}. $$

My notions are to use either $\mathbf{E}[X] = \int_{-\infty}^{\infty} xf(x) \, \mathrm{d}x$ where $f$ is the probability density function of $X$ and $\mathbf{P}(X \leq a) = \int_{-\infty}^{a} f(x) \, \mathrm{d}x$, or use the indicator method $\mathbf{E}[I_X] = \mathbf{P}(X)$ to substitute into the original inequality.
However, I got stuck midway in both approaches.
Thanks in advance for your help. I would highly appreciate it if a proof based on basic theorems of probability/expectation is provided.
 A: We have that $\max(X,a)\geq X$ and $\max(X,a)\geq a$, so taking expectation, $E(\max(X,a))\geq E(X)$ and $E(\max(X,a))\geq a$  Hence, $E(\max(X,a))\geq \max(E(X),a)$.
A: Here's an alternative approach: denote the probability space by $(\Omega,\mathcal{A},\mathbb{P})$ and use that $$\max\{X,a\} = \frac{X+a+|X-a|}{2} \tag{1}$$and that $\mathbb{E}$ is a linear operation together with $\mathbb{E}[a] = a$ to get $$\mathbb{E}[\max\{X,a\}] = \frac{\mathbb{E}[X]+a + \mathbb{E}[|X-a|]}{2}. \tag{2}$$Now observe that $$\begin{align}|\mathbb{E}[X]-a| &= \left|\int_\Omega X(\omega)\,{\rm d}\mathbb{P}(\omega) - a\right| \\ &= \left| \int_{\Omega} (X(\omega)-a)\,{\rm d}\mathbb{P}(\omega)\right| \\ &\leq \int_\Omega |X(\omega)-a|\,{\rm d}\mathbb{P}(\omega) \\ &= \mathbb{E}[|X-a|], \tag{3}\end{align}$$as $a = \int_\Omega a\,{\rm d}\mathbb{P}(\omega)$ and we use the triangle inequality for integrals. Plug (3) into (2) and use (1) again (with $\mathbb{E}[X]$ playing the role of $X$) to obtain $\mathbb{E}[\max\{X,a\}] \geq \max\{\mathbb{E}[X],a\}$.
