Is there a Markov-type inequality for the Median? Markov's theorem states that $P(|X| \geq a) \leq \frac{E[|X|]}{a}$. Is there an similar type of inequality that involves the median (somehow I doub't it, but I make no claim to comprehensive knowledge of probability inequalities).
What if we restrict $X$ to a non-negative, continuous pdf with a smooth, unimodal density function?

If $X$ is a random variable with finite variance $\sigma^2$, we have that the distance between the median and $\mathbb{E}[X]$ is at most $\sigma$ by Cantelli's inequality, hence the answer is affirmative under slightly stronger assumptions ($X\in L^2$ instead of just $X\in L^1$).
That assumptions gives, for instance:
$$\mathbb{P}[X>2\cdot\text{med}(X)]\leq \frac{\mathbb{E}[X]}{2\cdot\text{med}(X)}\leq\frac{1}{2}\cdot\min\left(\frac{\text{med}(X)+\sigma}{\text{med}(X)},\frac{\mathbb{E}[X]}{\mathbb{E}[X]-\sigma}\right).$$

On the other hand, $X\in L^2$ is somewhat a necessary assumption to work with the median. 
We may consider that $f_X(x)=\frac{3\sqrt{3}}{4\pi}\cdot\frac{1}{|x|^3+1}$ is the density of a random variable $X\in L^1\setminus L^2$.
For any $n>0$, we may define $X_n$ as the random variable with density:
$$ f_{X_n}(x)=\frac{3\sqrt{3}}{4\pi}\cdot\left\{\begin{array}{rcl}\frac{1}{|x|^3+1}&\text{if}& x<0,\\ \frac{n\,x^{n-1}}{x^{3n}+1}&\text{if}&x\geq 0.\end{array}\right.$$
Then the median of $X_n$ is always zero, but:
$$ \mathbb{E}[X_n] = -\frac{1}{2}+\frac{3\sqrt{3}}{4\pi}\int_{0}^{+\infty}\frac{n\,x^n}{1+x^{3n}}\,dx = -\frac{1}{2}+\frac{\pi}{3\sin\frac{\pi(n+1)}{3n}}$$
is unbounded as $n\to\left(\frac{1}{2}\right)^+$.
