Let $a = \frac12 \ln (2)$. Show that, for $x\ge 0.6$,
$$ \sinh(x) - \frac12 {\Large(} \frac {\cosh(x)}{\cosh(x-a)} {\Large)}^{x/a} > 0 $$
The LHS has a maximum at $x \simeq 1.72$ and is approaching zero for larger $x$. However, I find no clue how to show this, since also calculus methods give no closed results.
I also conjecture that this relation holds for other $a$ and for $x\ge x_0$, where $x_0$ will depend on $a$. See my edit in RiverLi's answer which shows that this is actually true for all $a>0$, and which shows $x_0(a)$.
With the substitution $x = \ln t$ (see @ablmf's answer), it suffices to prove that for all $t \ge \mathrm{e}^{3/5} \approx 1.8221$, $$\ln\Big(1 - \frac{1}{t^2}\Big) + \frac{2\ln t \ln (1 + \frac{1}{1+t^2})}{\ln 2} > 0.$$
We will use the following bounds (see 1): \begin{align} \ln (1 + x) &\ge \frac{x(6+5x)}{2(1+x)(3+x)}, \quad \forall x\in (-1, 0],\tag{1}\\ \ln (1+x) &\ge \frac{2x}{2+x}, \quad \forall x\ge 0.\tag{2} \end{align}
Applying (1) for $\ln (1-\frac{1}{t^2})$ and (2) for $\ln(1+\frac{1}{1+t^2})$, it suffices to prove that for all $t \ge \mathrm{e}^{3/5}$, $$\ln t - \frac{(6t^2-5)(2t^2+3)\ln 2}{8(t^2-1)(3t^2-1)} > 0.$$ Let $f(t) = \ln t - \frac{(6t^2-5)(2t^2+3)\ln 2}{8(t^2-1)(3t^2-1)}.$ We have $$f'(t) = \frac{18t^8 - (48 - 36\ln 2)t^6+(44 - 57\ln 2)t^4+(26\ln 2 - 16)t^2+2}{2t(3t^2-1)^2(t^2-1)^2}.$$ Let us prove that $f'(t) > 0$ for $t > 1$. It suffices to prove that for $t > 1$, $$g(t) = 18t^8 - (48 - 36\ln 2)t^6+(44 - 57\ln 2)t^4+(26\ln 2 - 16)t^2+2 > 0.$$ Let $t = 1 + u$ for $u > 0$. We have \begin{align} g(1+u) &= 18 u^8+144 u^7+(36 \ln 2+456) u^6+(216 \ln 2+720) u^5\\ &\quad +(483 \ln 2+584) u^4 +(492 \ln 2+224) u^3\\ &\quad +(224 \ln 2+32) u^2+ (40 \ln 2) u + 5 \ln 2. \end{align} Clearly, $g(1+u) > 0$. The desired result follows.
Thus, we have $f(t) \ge f(\mathrm{e}^{3/5}) \approx 0.000516 > 0$ for all $t \ge \mathrm{e}^{3/5}$. We are done.
Similarly, we may prove that for $a > 0$, the inequality holds for $x \ge x_0$ where $x_0$ depends on $a$.
EDIT by Andreas: This doesn't change RiverLi's answer above, but expands it for arbitrary $a > 0$. Going along the same paths as above, we obtain
$$ f(t) = \ln t - \frac{a (6t^2-5)(2t^2+1+e^{2a})}{4(t^2-1)(3t^2-1)(e^{2a}-1)} > 0. $$ The special case above for $a = \frac12 \ln (2)$ obviously follows. Now one can again examine $f'(t= 1+u)$ and obtains the necessary and sufficient condition for the numerator, \begin{align} g(1+u) &= (18 \exp(2 a) - 18) u^8 \\ &\quad+ (144 \exp(2 a) - 144) u^7 \\ &\quad+ (36 a + 456 \exp(2 a) + 18 a \exp(2 a) - 456)u^6 \\ &\quad+ (216 a + 720 \exp(2 a) + 108 a \exp(2 a) - 720) u^5 \\ &\quad + (486 a + 584 \exp(2 a) + 240 a \exp(2 a) - 584) u^4 \\ &\quad+ (504 a + 224 \exp(2 a) + 240 a \exp(2 a) - 224)u^3 \\ &\quad+ (240 a + 32 \exp(2 a) + 104 a \exp(2 a) - 32) u^2 \\ &\quad+ (48 a + 16 a \exp(2 a)) u \\ &\quad+ 6 a + 2 a \exp(2 a) \end{align} Since $\exp(2a)>1$, this is positive for all $u>0$. This means that the inequality holds at least for all $t>t_0(a)$ which satisfy $f(t_0) = 0$. Note again that this corresponds to $x>\ln (t_0(a))$. Since $f(t_0) = 0$ can only be solved implicitely, the results are visualized as follows:
The special case considered above may be identified in this figure: $a = \frac12 \ln 2 \simeq 0.347$, $t_0 < e^{3/5} \simeq 1.822$. The limiting cases are $t_0(a=0) \simeq 2.0355$, or $x_0(a=0) \simeq 0.71$, and since for large $t$, $f(t) \simeq \ln t - a/(\exp(2a) - 1) - \frac{1}{t^2}\frac{a(\exp(2a)/2 + 1)}{\exp(2a) - 1}$, we have that for large $a$, $t_0^2(a) \ln (t_0(a)) \simeq \frac{a}{2}$ which gives for large $a$, $t_0(a) <\simeq (\frac{e \; a}{2})^{1/3}$ or $x_0(a) <\simeq \frac13 \ln (\frac{e \; a}{2})$
1 Topsøe, "Some bounds for the logarithmic function", https://rgmia.org/papers/v7n2/pade.pdf