Let define the function :
$$f\left(x\right)=\frac{2}{x\left(\tanh\left(xe^{-1}\right)+1\right)}$$
Show that :
$$f\left(\frac{1+\sqrt{3}}{2}\right)<1$$
Some facts :
$$\tanh(x)=\sum_{n=1}^{\infty}\frac{2^{2n}(2^{2n-1}-1)B_{2n}z^{2n-1}}{(2n)!}$$
Where $B_{2n}$ ar the Bernoulli's numbers
The solution of :
$$f(x)=1$$
Can be find using the Lambert's function .
The function $f(x)$ is clearly decreasing on $(0,\infty)$
Question :
How to show the claim by hand without calculator ?
Some thoughts:
Note: $\tanh y = \frac{\mathrm{e}^y - \mathrm{e}^{-y}}{\mathrm{e}^y + \mathrm{e}^{-y}}$.
Denote $a = \frac{1 + \sqrt3}{2}$. It suffices to prove that $$a(\tanh(a\mathrm{e}^{-1}) + 1) > 2$$ or $$a\left(\frac{\mathrm{e}^{a \mathrm{e}^{-1}} - \mathrm{e}^{-a\mathrm{e}^{-1}}}{\mathrm{e}^{a\mathrm{e}^{-1}} + \mathrm{e}^{-a\mathrm{e}^{-1}}} + 1\right) > 2$$ or $$\mathrm{e}^{a \mathrm{e}^{-1}} > \frac{1}{\sqrt{a - 1}}$$ or $$\mathrm{e}^{-1} > \frac{1}{2a}\ln\frac{1}{a - 1}$$ or (using $\frac{1}{a - 1} = 2a$) $$\frac{\ln \mathrm{e}}{\mathrm{e}} > \frac{\ln(1 + \sqrt3)}{1 + \sqrt{3}}.$$
Let $f(x) = \frac{\ln x}{x}$. We have $f'(x) = \frac{1 - \ln x}{x^2}$. Thus, $f(x)$ is strictly decreasing on $[\mathrm{e}, \infty)$. It suffices to prove that $$\mathrm{e} < 1 + \sqrt3.$$