Is it possible for the median to be left to the mean in the Gamma distribution? Many textbooks teach a rule of thumb stating that the mean is right of the median under right skew, and left of the median under left skew but this is not always true.
According to wikipedia, there is no simple closed form for the median of the Gamma distribution.
Intuitively, I would think that the mean is always left to the median in the Gamma distribution. The mean would eventually be roughly equal to the median when the shape parameter $\alpha$ is very big. Is my intuition correct?

You can address this problem analytically.
Disregarding the rate parameter $\beta$ which cancels out, the mean is $\alpha$, and the median is such that the value of the CDF is $50\%$.
Plugging the mean value in the CDF and comparing to $50\%$ will tell you on what side of the median you are:
$$\frac{\gamma(\alpha,\alpha)}{\Gamma(\alpha)}\lessgtr\frac12.$$
A plot of this function leaves little doubt.

From the NIST library, we get the following asymptotic expansion for large $z$:
$$\mathop{\Gamma}\nolimits\!\left(z,z\right)\sim z^{z-1}e^{-z}\left(\sqrt{%
\frac{\pi}{2}}z^{\frac{1}{2}}-\frac{1}{3}+\frac{\sqrt{2\pi}}{24z^{\frac{1}{2}}%
}-\frac{4}{135z}+\frac{\sqrt{2\pi}}{576z^{\frac{3}{2}}}+\frac{8}{2835z^{2}}+%
\dots\right),$$
which confirms the trend to $\dfrac12$ when divided by $\Gamma(z)$. Showing a negative derivative (using 8.8.13 and 8.8.14 ?) would make it a proof of the original claim.
