# Best way to check to see if conditions are (or can be) satisfied?

Sorry to bother you guys, but a project I'm working on necessitates checking to see whether somewhat obscure conditions can all hold, given some bounds on possible values for each variable of interest. Specifically the one I'm working on now is whether: $$\frac{p_h-\bar{p}}{p_h-p_l} \leq \alpha^{\frac{(1-p_h)^{\alpha}-(1-p_l)^{\alpha}}{(1-p_h)^{\alpha}-(1-\bar{p})^{\alpha}}}$$ Can hold, given:

$0<p_l<\bar{p}<p_h<1$

and

$0<\alpha<1$

So, as it relates to this project, are all three conditions above satisfiable? That said, more broadly, do you guys have any suggestions on how to tackle problems like this in the future? (I'll have several more equations of a similar form, so any insights into the best way to approach problems like this will be greatly appreciated!!)

Sorry if this is a waste of time- a lot of the time I think I get a bit stuck on figuring out the best way to even start the process of testing conditions like this.

Thank you for the help!!

A solution does exist. The inequality $$0<p_l<\bar{p}<p_h<1$$ implies $$p_h-\bar{p}<p_h-p_l\Rightarrow \frac{p_h-\bar{p}}{p_h-p_l}<1$$
Hence $\alpha=1$ would work if it was allowed: $$\frac{p_h-\bar{p}}{p_h-p_l} \leq \alpha^{\frac{(1-p_h)^{\alpha}-(1-p_l)^{\alpha}}{(1-p_h)^{\alpha}-(1-\bar{p})^{\alpha}}}\overset{\alpha=1}{=}1$$
There is also a solution with $0<\alpha<1$, because $\alpha^{\frac{(1-p_h)^{\alpha}-(1-p_l)^{\alpha}}{(1-p_h)^{\alpha}-(1-\bar{p})^{\alpha}}}$ is continuous at $\alpha=1$ and the left hand side is strictly less than $1$.
Using software the answer can be found by a single 3D plot using an appropriate $\alpha$-value.
Feasible region for fixed $\alpha$:
With[{\[Alpha] = 2/3}, RegionPlot3D[0 < pl < pb < ph < 1 && Log[(ph - pb)/(ph - pl)] <= ((1 - ph)^\[Alpha] - (1 - pl)^\[Alpha])/((1 - ph)^\[Alpha] - (1 - pb)^\[Alpha]) Log[\[Alpha]], {ph, 0, 1}, {pb, 0, 1}, {pl, 0, 1}, MaxRecursion -> 0, PlotPoints -> 75]]