# Numerical results for Reduce on an inequality

I am trying to determine the regions parameters (a, b) must belong to for a function to be negative for any argument x greater than zero. I have tried using Resolve

Resolve[
ForAll[x, (a/x^2)*(Log[x - b] + 1)
- (a/x)*(1/(x - b)) -
(1/x^2)*(-b^3 + b) < 0] && x > 0, Reals]


but no expression could be provided. This is not surprising, as I suppose such expression would involve the solution of trascendental equations.

On the other hand, is there a way to get a numerical output from Resolve? For example, a set of (a, b) parameters values on the boundary of the region for which the function is positive?

I thought maybe Nsolve could take an inequality as an expression, but could not find any examples and could not make it work.

A "brute force" alternative I thought about is to create a grid of (a,b) values, and for each couple of values verify if the function is negative along all the positive x-axis (the last step does not seem obvious either). Is there a maybe more clever way, fully exploiting Mathematica's capabilities?

I got intrigued by the answer to this post, Reduce a complex inequality but I could not make it work for my case, for example

RegionPlot[
ForAll[x, (a/x^2)*(Log[x - b] + 1) - (a/x)*(1/(x - b)) - (1/
x^2)*(-b^3 + b) < 0], {a, 0.01, 0.5}, {b, 0.001,
0.01}]


does not seem to be correct syntax.

Thanks a lot

The region defined by the inequality is very complicated. It is unlikely that it can be described through any solve process.

cons = Simplify[(a/x^2)*(Log[x - b] + 1) - (a/x)*(1/(x - b)) - (1/x^2)*(-b^3 +
b) < 0, x > 0 && x > b] && x > 0 && x > b

(* a (-b + x) Log[-b + x] < b (a + (-1 + b^2) (b - x)) && x > 0 && x > b *)


The following plot is very slow

RegionPlot3D[cons,
{x, 0, 20}, {a, -30, 10}, {b, -25, 5},
AxesLabel ->
(Style[#, 14, Bold] & /@ {x, a, b}),
MaxRecursion -> 10,
PlotPoints -> 200,
PerformanceGoal -> "Quality"] // Quiet • thanks a lot very useful. – Smerdjakov Jan 10 at 22:16