Prove that $|r-\mu|\le \sigma\sqrt{2}$ Let $r$ be the median of a random variable $X$. Let $\mathbb{E}X = \mu <\infty$ and $D^2 (X)=\sigma^2<\infty$.
Prove that $|r-\mu|\le \sigma\sqrt{2}$.

Obviously $\mathbb{P}(X\ge r)=\mathbb{P}(X\le r)\ge\frac{1}{2}$.
Any hint please?
Edit 1:
So I have $\mathbb{P}(|X-\mu|\ge\sqrt{2}\sigma)\le\frac{1}{2}$ and $\mathbb{P}(|X-\mu|<\sqrt{2}\sigma)\ge\frac{1}{2}$. For $X=r$ the desired inequality holds.

Hint: Use the Chebyshev Inequality with $k=\sqrt{2}$.
