If $X$ follows exponential distribution with mean $\frac{1}{\theta}$, show that $E|X-a| \ge \frac{ln 2}{\theta}$ If $X$ follows exponential distribution with mean $\frac{1}{\theta}$, show that $E|X-a| \ge \frac{ln 2}{\theta}$ where $a$ is any real

I have this approach:
We know, median for exponential distribution is $\frac{\ln 2}{\theta}$
and mean deviation about median is least,
so, $E|X-a| \ge E|X-\frac{\ln 2}{\theta}| $

But how to show that it is greater than $\frac{\ln 2}{\theta}$?
Unfortunately, this argument isn't valid. Having $X \geq 0$ is not sufficient to claim that $E|X-a| \geq E|-a|$. Example: Let $X$ be $0$ or $1$, each with probability $1/2$, and let $a = 1$; then $E|X-a| = 1/2$, but $E|-a| = 1$. 

EDIT: My answer was downvoted, potentially because it was an answer to the original question but no longer to the edited question. As such, I'll edit to incorporate my comment below.

To compute median deviation about the mean, use the Law of the Unconscious Statistician:

$$\mathbb E \left[ \left| X - \frac{\ln 2}{\theta} \right| \right] = \int_0^{\frac{\ln 2}{\theta}} \theta e^{- \theta x} \cdot \left( \frac{\ln 2}{\theta} - x \right) \, \text d x + \int_{\frac{\ln 2}{\theta}}^{\infty} \theta e^{-\theta x} \cdot \left( x - \frac{\ln 2}{\theta} \right) \, \text d x$$
Unfortunately, this argument isn't valid. Having $X \geq 0$ is not sufficient to claim that $E|X-a| \geq E|-a|$. Example: Let $X$ be $0$ or $1$, each with probability $1/2$, and let $a = 1$; then $E|X-a| = 1/2$, but $E|-a| = 1$. 

EDIT: My answer was downvoted, potentially because it was an answer to the original question but no longer to the edited question. As such, I'll edit to incorporate my comment below.

To compute median deviation about the mean, use the Law of the Unconscious Statistician:

$$\mathbb E \left[ \left| X - \frac{\ln 2}{\theta} \right| \right] = \int_0^{\frac{\ln 2}{\theta}} \theta e^{- \theta x} \cdot \left( \frac{\ln 2}{\theta} - x \right) \, \text d x + \int_{\frac{\ln 2}{\theta}}^{\infty} \theta e^{-\theta x} \cdot \left( x - \frac{\ln 2}{\theta} \right) \, \text d x$$