I have Chebyshev’s inequality $$ P\left(\big|X-E[X]\big|\geq\epsilon\right)\leq\frac{V[X]}{\epsilon^2}, $$ where $E[X]$ is the expected value and $V[X]$ the variance.
Given I know that the median is $M$, $E[X]=\mu$ and $E[X^2]<\infty$, I have to show that: $$ |M-\mu|\leq\sqrt{2V[X]} $$
I thought about setting $\epsilon=\frac{1}{2}$, as I then will obtain the right side: $$ \sqrt{P\left(\big|X-\mu\big|\geq\frac{1}{2}\right)}\leq\sqrt{2V[X]}, $$ however, this leads to a blind end for me.
Then I thought about wanting to use the information about the median to set: $$ \frac{V[X]}{\epsilon^2}=\frac{1}{2}\quad\quad\Rightarrow\quad\quad \epsilon=\sqrt{2V[X]}, $$ and I will then get: $$ P\left(\big|X-\mu\big|\geq\sqrt{2V[X]}\right)\leq\frac{1}{2}, $$ where I thought I could use the median, when I have $\frac{1}{2}$ on the right side. However, I'm still quite stuck.
Any help regarding this inequality will be greatly appreciated.