Logarithmic convexity of incomplete gamma function. The incomplete Gamma function $F(t)$ satisfies
$$
1 - F(t) \sim \int^\infty_t dx \, e^{-x} x^{\alpha - 1}
$$
for $t > 0$ and its derivative $F^\prime \sim e^{-t} t^{\alpha - 1}$ is the density function of the Gamma distribution.
It is claimed that the hazard rate, also known as the failure rate
$$h = -\frac{d}{dt} \ln (1 - F) = \frac{F^\prime}{1-F}$$
is increasing everywhere (i.e. $h^\prime (t) > 0$) for the case that $\alpha > 1$ and decreasing everywhere for the case that $\alpha < 1$.
In other words, the logarithmic convexity of $1 - F$ depends only on the sign of $\alpha - 1$.
This result is also given as a proof exercise in problem 5.22 of this book.
It's unclear to me how to prove this result. We have
$$ h^\prime = \frac{(1-F)F^{\prime \prime} + (F^\prime)^2}{(1-F)^2} = \frac{F^\prime}{(1-F)^2} \left[ (1-F)\left(\frac{\alpha - 1}{t} - 1 \right) + F^\prime \right], $$
but from this point the next step is unclear.

Consider $t=0$. Then, $h(0)=0$ and since for any $\epsilon > 0$ we have $h(\epsilon) > 0$, the derivative (to the right) at zero is positive. (I assume we can omit to consider $t<0$).
Now, consider $t > 0$. On this set, $1/h$ is well defined and given by
$$1/h = \int_t^{\infty} dx e^{t-x} \left(\frac{x}{t}\right)^{\alpha - 1}$$
Doing a change of variable, we get
$$1/h = \int_0^{\infty} dx e^{-x} \left(1+\frac{x}{t}\right)^{\alpha - 1}$$
Now, note that the term
$$\left(1+\frac{x}{t}\right)^{\alpha - 1}$$
increases (decreases) in $t$ if $\alpha<1$ ($\alpha>1$).
The claim then follows.
