Proving that a function is monotone Here is the setting: We have a middleman that buys a product from the producers, and sells the product to the customers. The middleman charges a price $R$ to the customers, and pays a price $p(R)$ to buy the product from the producers.
We use $G(R)$ to denote a cumulative density function for a probability distribution, with $G(0)=0$ and $G(\infty)=1$. The $1-G(x)$ distribution represents the customer's willingness to pay for a product, and it is the fraction of consumer that will buy the product if it is priced at $R$.
We know (from other analysis) that the optimal price to pay to the producers of the product $p(R) = R-\frac{1-G(R)}{G'(R)}$. (Assume that the middleman is a monopsonist with the power to set the price of buying.)
Assuming that we are interested only in regions where $p(R)<R$ (i.e., we make a profit when selling) and $p(R)>0$ (i.e., we need to pay a non-zero price to procur the product from the producer), can we prove that the function $p(R)$ does not decrease when we increase the price $R$? In other words, if we increase the price R that we charge customers, can we show that the price that we are willing to pay the supplier also increases?
By setting $\frac{d}{dR}p(R) \geq 0$ we get $G''(R) \geq -2 \cdot \frac{\left(G'(R)\right)^2}{1-G(R)}$
Can we prove that the latter holds, assuming that G(R) can be an arbitrary probability distribution with support in $(0..\infty)$ and that we only care for the regions where $p(R)<R$ and $p(R)>0$? Assume that G'(R) and G''(R) exist, that they are continuous, differentiable, etc.
 A: So in general I know that this does not hold, because your expression is exactly the myerson virtual value, and there is a long literature about suitable regularity conditions to make it increasing. If you assume log-concavity of your distribution for example then you are done. But this is too strong in that you can actually assume -.5 concavity. I refer you to optimal design and $\rho$-concavity.
http://citeseerx.ist.psu.edu/viewdoc/download?
doi=10.1.1.362.7589&rep=rep1&type=pdf
A: OK, it seems that I managed to solve this.
Taking the derivative of p(R) and by setting it to be greater than or equal to 0, we have:
$G′′(R)⋅(1−G(R))≥−2⋅G′(R)⋅G′(R)$
Taking the integral of both sides:
$∫G′′(R)−∫G′′(R)⋅G(R)≥−2∫G′(R)⋅G′(R)$
Taking integration by parts for $∫G′′(R)⋅G(R)$ we have:
$G′(R)−G(R)⋅G′(R)≥−3∫G′(R)⋅G′(R)$
Taking the integral of both sides again:
$G(R)−∫G(R)⋅G′(R)≥−3∫∫G′(R)⋅G′(R)$
We have that ∫G(R)⋅G′(R)=12G(R)2 and this results in:
$G(R)(1−\frac{1}{2}G(R))+3∫∫G′(R)⋅G′(R)≥0$
Since $G(R)≥0$, $1−\frac{1}{2}G(R)≥0.5$ and $G′(R)≥0$, we have a sum of positive terms, which is positive. QED
