Probability - median and mean

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I'm having a difficult time with this one. Would be happy for some help.

I'm kind of gets why its true but can't manage to prove it.

Let $X$ be a random variable with median $\Bbb MX$ such that positive constants $a, b$ exist so that for all $t > 0$, $P(|X − \Bbb MX| > t) ≤ ae^{−t^2/b}$.

Show that $|\Bbb MX−\Bbb EX|≤min(\sqrt{ab},a\sqrt{b\pi}/2)$.

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For one upper bound I don't think you have to use the fact that you are dealing with the median:

If you define $Y:=\vert X-\mathbb{M}(X)\vert$, then $\mathbb{E}[Y]\geq \vert \mathbb{E}[X]-\mathbb{M}(X)\vert$ by the triangle inequality for expectations. Using the continuous tail-sum formula,

$$ \mathbb{E}[Y]=\int _0^\infty\mathbb{P}(Y>t)dt, $$

and the assumed inequality we get that $\mathbb{E}[Y]\leq \int_0^\infty ae^{-\frac{t^2}{b}}dt= \frac{a\sqrt{b\pi} }{\sqrt{2}}$.

I don't know how to obtain the other bound, but maybe this will help.