Why is $x(1-F(x))$ a concave function when the function $F(x)$ is a monotone hazard rate distribution? It is not clear to me why $x(1-F(x))$ is concave when $F(x)$ is a monotone hazard rate distribution.
This fact is mentioned in [1] and [2].
I understood that $x(1-F(x))$ is unimodal from reference [3].
Such descriptions are in the second paragraph of section 3.1.1 of [1] and the first paragraph of section 3 of [2].
Best regards.
[1] Tong, Yongxin, et al. "Dynamic pricing in spatial crowdsourcing: A matching-based approach." Proceedings of the 2018 International Conference on Management of Data. 2018.
https://tik-old.ee.ethz.ch/file//935269ac46a7de0505d43ae598542922/sigmod18-tong.pdf
[2]Babaioff, Moshe, et al. "Dynamic pricing with limited supply." ACM Transactions on Economics and Computation (TEAC) 3.1 (2015): 1-26.
https://arxiv.org/abs/1108.4142
[3]Regular and MHR Distributions (Web site)
http://cs.brown.edu/courses/cs1951k/lectures/2020/mhr_regular_dist.pdf
 A: What is going on is that the authors require $1-F(x)$ to be log concave.  Sorry, the answer will start slowly with a somewhat trivial problem, but it will illustrate what is going on and then give a more direct answer at the end.
The intuition comes from the monopoly problem.  If a consumer's value is $v$, he only buys at a price of $t$ if $ v \ge t$.  If $v$ is distributed $F$, that means the probability of a purchase given a price of $t$ is $1-F(t)$.  The monopolist then sets a price (assuming a cost of zero to keep things simple) to solve
$$
\max_t t(1-F(t)).
$$
The FONC is
$$
1-F(t) - tf(t) = 0
$$
and the SOSC is
$$
-2f(t) - tf'(t) <0.
$$
That SOSC is not very useful.  We don't generally like to impose assumptions on derivatives of densities, because densities are already somewhat "poorly behaved": we don't have many a priori reasons to assume they take one shape or another.  
What would be more useful, though, are conditions under which the FONC is a decreasing function in $t$, so that any critical point is a global maximizer.  Let's go back to that and re-arrange:
$$
\phi(t) = \dfrac{1-F(t)}{f(t)} - t
$$
The $-t$ part is clearly decreasing.  If we assume that $(1-F(t))/f(t)$ is decreasing, then $\phi(t)$ is decreasing. That quantity is decreasing if
$$
\dfrac{f(t)}{1-F(t)}
$$
is increasing.  This is just the hazard rate, and notice that
$$
D^2(1-F(t)) = - D \dfrac{f(t)}{1-F(t)}.
$$
So if $f/(1-F)$ is increasing, $D^2(1-F(t))$ is decreasing, and $1-F(t)$ is log-concave.
So this whole game is about ensuring that $t(1-F(t))$ is a concave function so that FONCs are sufficient.  A simple condition for that is that $\log(t) + \log(1-F(t))$ is concave.  Or that $1-F(t)$ is log-concave (which is a direct but not very illuminating answer).
A good reference is Bergstrom and Bagnoli, Economic Theory, Log-Concave Probability and its Applications.
