Remark: Although we get a closed-form bound (in terms of elementary functions), the solution is still quite complicated.
We have
$$\int_{-\infty}^0 (\cosh x - 1)^x\, \mathrm{d}x
= \int_0^\infty 2^{-x}\left(\sinh\frac{x}{2}\right)^{-2x}\,\mathrm{d} x = \int_0^\infty 2^{1 - 2x}(\sinh x)^{-4x}\, \mathrm{d}x.$$
Fact 1: $\int (\sinh x)^{-5}\, \mathrm{d}x =
-\frac{\cosh x}{4\sinh^4 x} + \frac{3\cosh x}{8\sinh^2 x} - \frac{3}{8}\ln\frac{\mathrm{e}^x + 1}{\mathrm{e}^x - 1} + C$.
Fact 2: $\int (\sinh x)^{-4}\, \mathrm{d}x =
-\frac{\cosh x}{3\sinh^3 x} + \frac{2\cosh x}{3\sinh x} + C$.
Fact 3: $\sinh x \ge x + \frac16 x^3$ for all $x\in [0, 1]$.
Let $T(f(x),a,n)$ denote the $n$-th order Taylor approximation of $f(x)$ at $x = a$.
Fact 4: For all $x\in [0, 1]$,
$(1 + x^2/6)^{-4x} \le T((1 + x^2/6)^{-4x}, 0, 12)$.
(Proof: Take logarithm, then take derivative and the 2nd derivative.)
Fact 5: For all $x\in [0, 1]$,
$2^{-2x} \le T(2^{-2x}, 0, 6)$.
(Proof: Take logarithm, then take derivative.)
Fact 6: For all $u \in [-1, 0]$,
$\mathrm{e}^{-4u} \le T(\mathrm{e}^{-4u}, 0, 8) + \frac{512}{315}u^8$.
(Proof: Take logarithm, then take derivative.)
First, using Fact 1, we have
\begin{align*}
\int_{5/4}^\infty 2^{1 - 2x}(\sinh x)^{-4x}\, \mathrm{d}x
&\le \int_{5/4}^\infty 2^{1 - 2\cdot \frac{5}{4}}(\sinh x)^{-4\cdot \frac{5}{4}}\, \mathrm{d}x \\
&= -\frac{\sqrt2}{4}\left(-\frac{\cosh (5/4)}{4\sinh^4 (5/4)} + \frac{3\cosh (5/4)}{8\sinh^2 (5/4)} - \frac{3}{8}\ln\frac{\mathrm{e}^{5/4} + 1}{\mathrm{e}^{5/4} - 1}\right)\\
&< \frac{3}{500}.
\end{align*}
Second, using Fact 2, we have
\begin{align*}
\int_1^{5/4} 2^{1 - 2x}(\sinh x)^{-4x}\, \mathrm{d}x
&\le \int_1^{5/4} 2^{1 - 2\cdot 1}(\sinh x)^{-4\cdot 1}\, \mathrm{d}x\\
&= -\frac{\cosh (5/4)}{6\sinh^3 (5/4)} + \frac{\cosh (5/4)}{3\sinh (5/4)} - \left(-\frac{\cosh 1}{6\sinh^3 1} + \frac{\cosh 1}{3\sinh 1}\right)\\
&< \frac{3}{80}.
\end{align*}
Third, using Facts 3-6, we have
\begin{align*}
&\int_0^1 2^{1 - 2x}(\sinh x)^{-4x}\, \mathrm{d}x\\
\le\,& \int_0^1 2^{1 - 2x}(x + x^3/6)^{-4x}\, \mathrm{d}x\\
=\, & \int_0^1 2^{1 - 2x}(1 + x^2/6)^{-4x}x^{-4x}\, \mathrm{d}x\\
\le\, & 2\int_0^1 T(2^{-2x}, 0, 6) \, T((1 + x^2/6)^{-4x}, 0, 12)
\, \left[T(\mathrm{e}^{-4u}, 0, 8) + \frac{512}{315}u^8\right]_{u = x\ln x}\, \mathrm{d} x \tag{1}\\
<\, & 1549/500.
\end{align*}
Note: The integral in (1) admits a closed form
$a_0 + a_1\ln 2 + a_2\ln^2 2 + a_3\ln^3 2 + a_4\ln^4 2 + a_5\ln^5 2 + a_6\ln^6 2$
where $a_i$'s are all rational numbers.
Thus, we have $\int_0^\infty 2^{1 - 2x}(\sinh x)^{-4x}\, \mathrm{d}x
< \frac{3}{500} + \frac{3}{80} + 1549/500 < \pi$.
We are done.
Sketch of a proof:
Denote
$$A = (1 - x)^{2(1 - x)}, \quad B = (1 - x)^{2x - 1},
\quad C = x^{2x}.$$
The desired inequality is written as
$$\left(\frac{A}{(1 - x)^2} - \frac{B}{2^4x(1 - x)^2C}\right)(x^{-2}C - 1) \ge 1.$$
Denote $a = \ln 2$. Let
\begin{align*}
A_1 &= \frac{p_1x^2 + p_2x + p_3}{(8a^3 - 24a^2
+ 48a - 24) x - 4a^3 - 12}, \\
B_1 &= \frac{(2a -2)x^2 + (-3a + 2)x + a}{(4a - 2)x^2 + (-4a + 2)x + a}, \\
C_1 &= \frac{(4a - 2)x^2 + (-4a + 2)x + a}{(2a - 2)x - a + 2},
\end{align*}
where
\begin{align*}
p_1 &= -4a^4 + 16a^3 -24a^2 + 24a - 24, \\
p_2 &= 4a^4 - 24a^3 + 48a^2 - 48a + 36, \\
p_3 &= - a^4 + 8a^3 - 24a^2 + 30a - 24.
\end{align*}
Fact 1: $A \ge A_1 > 0$ for all $x \in [1/2, 1)$.
Fact 2: $B \le B_1$ for all $x \in [1/2, 1)$.
Fact 3: $C \ge C_1 > 0$ for all $x \in [1/2, 1)$.
Fact 4: $x^{-2}C_1 - 1 > 0$ for all $x \in [1/2, 1)$.
By Facts 1-4, it suffices to prove that
$$\left(\frac{A_1}{(1 - x)^2} - \frac{B_1}{2^4x(1 - x)^2C_1}\right)(x^{-2}C_1 - 1) \ge 1$$
which is true (simply a polynomial inequality).
We are done.
Best Answer
A piece of sheer numerical good luck makes the calculation fairly painless.
I'll take this inequality to be well known: $$ \pi > \frac{333}{106}. $$ (I don't know a simple proof. For some that aren't simple, see Is there an integral that proves $\pi > 333/106$?.)
From $e$ Continued Fraction --- from Wolfram MathWorld, we have: $$ e < \frac{193}{71}. $$
Therefore, it is enough to prove: $$ \sqrt{\frac{193}{2\cdot71}} < \frac{333}{106}\left(\frac{193^2}{2\cdot71^2} - 1\right)^{-1}. $$ But \begin{multline*} 193^2 - 2\cdot71^2 = (200 - 7)^2 - 2(70 + 1)^2 = (40000 - 2800 + 49) - 2(4900 + 140 + 1) \\ = 37249 - 10082 = 27167 = 7\cdot3881, \end{multline*} and the right hand side of the required inequality simplifies to: $$ \frac{2\cdot3\cdot111\cdot71^2}{2\cdot7\cdot53\cdot3881} = \frac{426\cdot7881}{371\cdot7762} > \frac{426\cdot7882}{371\cdot7763} = \frac{426\cdot1126}{371\cdot1109} = \frac{426}{371}\left(1 + \frac{17}{1109}\right), $$ where we have used the fact that if $y > x > 0$ then $\frac{y}{x} > \frac{y + 1}{x + 1}.$
Now comes the piece of luck: $$ \frac{1109}{17} = 65 + \frac4{17} < 65 + \frac14 = \frac{261}4, \text{ whence } 1 + \frac{17}{1109} > 1 + \frac4{261}, $$ whence the right hand side of the required inequality is greater than: $$ \frac{426\cdot265}{371\cdot261} = \frac{(6\cdot71)\cdot(5\cdot53)}{(7\cdot53)\cdot(3\cdot87)} = \frac{710}{609}. $$ We now only have to prove: $$ \frac{193}{2\cdot71} < \left(\frac{710}{609}\right)^2, \text{ i.e., }\ 200\cdot71^3 > 193\cdot609^2. $$ We have already calculated $71^2 = 5{,}041,$ so $71^3 = 352{,}870 + 5{,}041 = 357{,}911,$ and $200\cdot71^3 = 71{,}582{,}200.$ On the other hand, $609^2 = 360{,}000 + 10{,}800 + 81 = 370{,}881,$ therefore $193\cdot609^2 = (200 - 7)\cdot370{,}881 = 74{,}176{,}200 - 2{,}596{,}167 = 71{,}580{,}033.$ This proves the inequality.