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 $$\sin(\tanh x) \ge \tanh(\sin x), ~~ \text{for}~~ x \in [0,\pi/2]$$ Interestingly, the first three terms of the Mclaurin series are identical for both the functions.
Here is a proof:
Fact 1: It holds that, for all $x \ge 0$, $$\tanh x \ge \frac{10x^3 + 105x}{x^4 + 45x^2 + 105}.$$ (The proof is given at the end.)
Fact 2: It holds that, for all $u \in [0, 1]$, $$\tanh u \le \frac{u^3 + 15u}{6u^2 + 15}.$$ (The proof is given at the end.)
Using Fact 1 and $\tanh x \le 1$, we have $$\sin (\tanh x) \ge \sin \left(\frac{10x^3 + 105x}{x^4 + 45x^2 + 105}\right). \tag{1}$$
Using Fact 2 and $\sin x \in [0, 1]$, we have $$\tanh(\sin x) \le \frac{\sin^3 x + 15\sin x}{6\sin^2 x + 15}. \tag{2}$$
Using (1) and (2), it suffices to prove that $$\sin \left(\frac{10x^3 + 105x}{x^4 + 45x^2 + 105}\right) \ge \frac{\sin^3 x + 15\sin x}{6\sin^2 x + 15}. \tag{3}$$
Let $$A := \frac{10x^3 + 105x}{x^4 + 45x^2 + 105}, \quad B := x - \frac{x^3}{3!} + \frac{x^5}{5!}.$$
Using $\sin u \ge u - \frac{u^3}{3!} + \frac{u^5}{5!} - \frac{u^7}{7!}$ for all $u\ge 0$, we have $$\sin \left(\frac{10x^3 + 105x}{x^4 + 45x^2 + 105}\right) \ge A - \frac{A^3}{3!} + \frac{A^5}{5!} - \frac{A^7}{7!}. \tag{4}$$
Using $\sin u \le u - \frac{u^3}{3!} + \frac{u^5}{5!}$ for all $u \ge 0$, noting that $u\mapsto \frac{u^3 + 15u}{6u^2 + 15}$ is strictly increasing on $u\ge 0$, we have $$\frac{\sin^3 x + 15\sin x}{6\sin^2 x + 15} \le \frac{B^3 + 15B}{6B^2 + 15}. \tag{5}$$
Using (3), (4) and (5), it suffices to prove that $$A - \frac{A^3}{3!} + \frac{A^5}{5!} - \frac{A^7}{7!} \ge \frac{B^3 + 15B}{6B^2 + 15}$$ or $$\frac{x^7 h(x)}{( {x}^{4}+45\,{x}^{2}+105) ^{7} ( {x}^{10}-40{x}^{ 8}+640{x}^{6}-4800{x}^{4}+14400{x}^{2}+36000) }\ge 0 \tag{6}$$ where \begin{align*} h(x) &:= -7\,{x}^{36}-1785\,{x}^{34}-181440\,{x}^{32}-8778175\,{x}^{30}- 152703075\,{x}^{28}\\ &\qquad +3298464750\,{x}^{26}+165337839125\,{x}^{24}+ 826367130625\,{x}^{22}\\ &\qquad -45036632734375\,{x}^{20}-480887233668750\,{x}^{ 18}+2020497175546875\,{x}^{16}\\ &\qquad +5494328646890625\,{x}^{14}+ 335334922673625000\,{x}^{12}\\ &\qquad +6005165669281406250\,{x}^{10} + 43232877600909609375\,{x}^{8}\\ &\qquad +170152296290845312500\,{x}^{6} + 390099811698253125000\,{x}^{4}\\ &\qquad +490539329059500000000\,{x}^{2}+ 262396086816937500000. \end{align*}
We can prove that $h(x) \ge 0$ on $[0, \pi/2]$. Thus, (6) is true.
We are done.
$\phantom{2}$
Proof of Fact 1:
Using $\tanh x = 1 - \frac{2}{\mathrm{e}^{2x} + 1}$, it suffices to prove that $$f(x) := 2x - \ln \frac{x^4 + 10x^3 + 45x^2 + 105x + 105}{x^4 - 10x^3 + 45x^2 - 105x + 105} \ge 0.$$
We have $$f'(x) = \frac{2x^8}{x^8 - 10x^6 + 135x^4 - 1575x^2 + 11025} \ge 0.$$ Also, $f(0) = 0$. Thus, we have $f(x) \ge 0$ for all $x \ge 0$.
We are done.
Proof of Fact 2:
Using $\tanh u = 1 - \frac{2}{\mathrm{e}^{2u} + 1}$, it suffices to prove that $$g(u) := \ln\frac{u^3 + 6u^2 + 15u + 15}{-u^3 + 6u^2 - 15u + 15} - 2u \ge 0.$$
We have $$g'(u) = \frac{2u^6}{-u^6 + 6u^4 - 45u^2 + 225} \ge 0.$$ Also, $g(0) = 0$. Thus, we have $g(u) \ge 0$ on $[0, 1]$.
We are done.
This is an attach from the right side of the interval, so only "half" of an argument. An attack from the left side would complete an argument. (Taylor's approximation might work, but it appears that you would need to establish an upper bound on the 8th derivative of $\sin(\tanh(x))-\tanh(\sin(x))$, which is not pretty.)
If $x$ is in $(0,\pi/2)$ and greater than $\tanh^{-1}(\arcsin(\tanh(\sin(\pi/2))))=\tanh^{-1}(\arcsin(\tanh(1)))$, then $\sin(\tanh(x))>\tanh(1)$, but $\tanh(\sin(x))$ is always at most $\tanh(1)$. So the claim is true on $\left(\tanh^{-1}(\arcsin(\tanh(1))),\frac{\pi}4\right)$.
Let $a_1=\tanh^{-1}(\arcsin(\tanh(\sin(\pi/2))))\approx1.316$.
Let $a_2=\tanh^{-1}(\arcsin(\tanh(\sin(a_1))))\approx1.237$.
Consider $x$ in $(a_2,a_1]$. Now $\sin(\tanh(x))>\tanh(\sin(a_1))$, but $\tanh(\sin(x))$ is at most $\tanh(\sin(a_1))$. So the claim is true on $\left(a_2,\frac{\pi}4\right)$.
In this manner, if we recursively define $a_n=\tanh^{-1}(\arcsin(\tanh(\sin(a_{n-1}))))$, we can show that the claim is true on $\left(a_n,\frac{\pi}4\right)$. And in fact, that it is true on $\left(\lim\limits_{n\to\infty}a_n,\frac{\pi}4\right)$.
So what is $L=\lim\limits_{n\to\infty}a_n$? If it is $0$, we are done. Well, we'd have to solve $\tanh^{-1}(\arcsin(\tanh(\sin(L))))=L$, equivalent to $\tanh(\sin(L))=\sin(\tanh(L))$. But if we were able to establish that the only solution that equation is $0$, then we'd have an easier proof in the first place (since our functions are continuous and it's easy to show that for very small $x$, $\sin(\tanh(x))>\tanh(\sin(x))$.
But numerically, $a_{12}<1$. (So says my computer. Tedious application of some rigorously bounded approximations of these functions could establish that.) So the claim holds at least on $[1,\pi/2]$. It would remain to show that the claim is true for $x$ on the left, that is, $x$ in $[0,1)$.
