I am faced with the following inequality:
$$3\sqrt{3}\cdot\sqrt[3]{4}^h > 4\pi h^2$$
After running it through Wolfram, I get the following inequality:
$$x>-\frac{3W_{-1}\big(-\frac{ln(2)}{2\sqrt[4]{3}\sqrt{\pi}}\big)}{ln(2)}$$
I'm not at all familiar with the Lambert-$W$ function, so I would appreciate some sort of explanation as to what this result means. I am mainly concerned about what this means about the possible values of $x$ (i.e. can the inequality be satisfied with $x\in \mathbb{R}$?).
Without Lambert function.
Consider the problem as $$\frac{3 \sqrt{3}\, 2^{2 h/3} }{4\pi h^2} >1$$ and taking logarithms $$\log\left(\frac{3 \sqrt{3}\, 2^{2 h/3} }{4\pi h^2} \right)>0$$ Expanding the logarithm (assuming $h>0$, consider the function $$f(h)=\log\left(\frac{3 \sqrt{3} }{4\pi } \right)+\frac{2h}3\log(2)-2\log(h)$$ and its derivatives $$f'(h)=\frac{2}3\log(2)-\frac 2h\qquad \qquad f''(h)=\frac 2{h^2}>0 $$ The first derivative cancels for $$h_*=\frac{3}{\log (2)}\approx 4.32809\implies f(h_*)=\log \left(\frac{2^{\frac{2}{\log (2)}-2} \log ^2(2)}{\sqrt{3} \pi }\right)\approx -1.81336$$ and then there are two values of $h$ which make $f(h)=0$.
We can compute these roots using Newton method. The iterates would then be
$$\left( \begin{array}{cc} n & h_n \\ 0 & 1 \\ 1 & 0.726245 \\ 2 & 0.766487 \\ 3 & 0.767866 \\ 4 & 0.767868 \end{array} \right)$$ then $h_1\approx 0.767868$.
$$\left( \begin{array}{cc} n & h_n \\ 0 & 10 \\ 1 & 13.3090 \\ 2 & 13.0201 \\ 3 & 13.0186 \end{array} \right)$$ then $h_2\approx 13.0186$.
So $f(h)<0$ if $h_1< h < h_2$.
With Lambert function.
What Wolfram Alpha gave you as a result is the analytical expression of $h_2$ but, if you plot the function, you will notice that one part is missing $(0 < h < h_1)$.
In fact, the equation has three roots which write $$h_0=-\frac{3 W\left(\frac{\log (2)}{2 \sqrt[4]{3} \sqrt{\pi }}\right)}{\log (2)}\approx -0.564418$$ $$h_1=-\frac{3 W\left(-\frac{\log (2)}{2 \sqrt[4]{3} \sqrt{\pi }}\right)}{\log (2)}\approx 0.767868$$ $$h_2=-\frac{3 W_{-1}\left(-\frac{\log (2)}{2 \sqrt[4]{3} \sqrt{\pi }}\right)}{\log (2)}\approx 13.0186$$ and for $h_0 < h < h_1$, $f(h) >0$ also.
Edit
Considering the small portion around $h=0$, you could notice that, using Taylor series for the orginal expression, we would get $$3\sqrt{3}\cdot\sqrt[3]{4}^h - 4\pi h^2=3 \sqrt{3}+2 \sqrt{3} \log (2)\,h)+ \left(\frac{2 \log ^2(2)}{\sqrt{3}}-4 \pi \right)h^2+O\left(h^3\right)$$ Ignoring the higher order terms, the roots of the quadratic are $$\frac{ 3 \left(\pm\sqrt{12 \sqrt{3} \pi -3 \log ^2(2)}+\sqrt{3} \log (2)\right)}{2 \left(6 \pi -\sqrt{3} \log ^2(2)\right)}$$ that is to say $\approx -0.565320$ and $\approx 0.765221$ to be compared with the exact values of $h_0$ and $h_1$. Between these two roots, the function is positive.
If, using Wolfram Alpha, you type
it gives you a range in terms of two Lambert functions