Proving $\sin(\tanh x) \ge \tanh(\sin x)$, for $x \in [0,\pi/2]$

Earlier, a very interesting proof of an inequality has been proposed at MSE:How prove this inequality $\tan{(\sin{x})}>\sin{(\tan{x})}$

Here the question is: How to prove that $$\sin(\tanh x) \ge \tanh(\sin x), ~~ for~~ x \in [0,\pi/2].$$ Interestingly, the first three terms of the Mclaurin series are identical for both the functions.

from Hot Weekly Questions - Mathematics Stack Exchange