Firstly,
$$\frac12\Gamma\left(\frac12\right)=\dfrac{\sqrt\pi}2\approx 0.886.\tag1$$
Denote
$$y=\frac12\Gamma\left(\frac12\Gamma\left(\frac12\right)\right),\tag2$$
then
$$y = \frac12\Gamma\left(\dfrac{1-x}{1+x}\right) < \frac12 + \gamma x + \left(\frac{\pi^2}6+\gamma^2-\gamma\right)x^2 + 0.251 x^3 + 2.33 x^4\tag3$$
(see also Wolfram Alpha series and coefficients),
where $\gamma$ is the Euler-Mascheroni constant,
$$x=\dfrac{2-\sqrt\pi}{2+\sqrt\pi}\approx0.060317809,\quad
\gamma\approx0.577215665.\tag4$$
The given inequality can be written in the form of
$$\dfrac1{\sqrt\pi}\Gamma(y) < \dfrac{\pi\sqrt\pi}6\approx0.928054666,\tag5$$
wherein
$$\dfrac1{\sqrt\pi}\Gamma(y) < 1 + \psi\left(\frac12\right)\left(y - \dfrac12\right) + \frac14\left(\pi^2+\psi^2\left(\frac12\right)\right)\left(y - \dfrac12\right)^2,\tag6$$
(see also Wolfram alpha series and coefficients),
where $\psi(x)$ is digamma function,
$$\psi\left(\frac12\right) = -\gamma - 2\ln2 \approx-1.963510026.\tag7$$
From $(3),(4)$ should
$$y-\dfrac12 < \frac1{25}.\tag8$$
From $(6)-(8)$ should
$$\dfrac1{\sqrt\pi}\Gamma(y) < 1 + \dfrac1{25}\psi\left(\frac12\right) + \dfrac1{2500}\left(\pi^2+\psi^2\left(\frac12\right)\right)\approx0.926949589
< \dfrac{\pi\sqrt\pi}6,$$
i.e. inequality $(5)$ is correct.
Therefore,
$$\color{brown}{\mathbf{\Gamma\left(\frac12\Gamma\left(\frac12\Gamma\left(\frac12 \right)\right)\right) < \frac{\pi^2}6}}.$$
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.
Best Answer
You may show the inequality using elementary equivalent rearrangements and Cauchy-Schwarz inequality as shown below.
This is surely not the nicest way but it works:
\begin{eqnarray*} \frac{1}{2\ln2}\left(1-\sqrt{\frac{1-\ln2}{1+\ln2}}\right) & > & \sqrt{2}-1 \\ & \Leftrightarrow & \\ \frac{1}{2\ln2}\left(1-\frac{1-\ln2}{1+\ln2}\right) & > & \left(\sqrt{2}-1\right)\left(1+\sqrt{\frac{1-\ln2}{1+\ln2}}\right) \\ & \stackrel{\frac 1{\sqrt 2 - 1} = \sqrt 2 + 1}{\Longleftrightarrow} & \\ \sqrt 2 + 1 & > & \left(1+\ln 2\right)\left(1+\sqrt{\frac{1-\ln2}{1+\ln2}}\right) \\ & \Leftrightarrow & \\ \sqrt 2 & > &\ln 2 + \sqrt{1-\ln^2 2} \\ \end{eqnarray*}
The last inequality is true because of Cauchy-Schwarz:
$$1\cdot \ln 2 + 1\cdot \sqrt{1-\ln^2 2} \stackrel{C.S.}{<}\sqrt 2 \cdot \sqrt{\ln^2 2 + 1 - \ln^2 2} = \sqrt 2$$