I'm trying to get Mathematica to solve a particular inequality in terms of the variables (and constants) of interest to find where the inequality holds as true. Specifically, I have three functions of interest:

b[a_, h_, l_]:= a*l + (1-a)*h
condition[f_, a_, h_, b_, l_]:= (((f[1-b[a,h,l]]-f[1-h])/(h-b[a,h,l]))/((f[1-l]-f[1-h])/(h-l)))-(f[a]/a)

In addition to these three functions, there are several conditions that will always apply for these variables:


Ideally, I was hoping for Mathematica to tell me whether the "condition" equation is every greater than 0 over the aforementioned ranges, and if so, for it to return the necessary conditions for it in terms of a, h, l (and/or b).

Unfortunately, I'm not sure what command to use, or whether the problem as specified above even makes any sense. Any insights that anyone has for how to solve this would be greatly appreciated!!

*Note that the function for f(p) will change, but always satisfy several requirements, such as: $f(0)=0$

$f(1) = 1$

$f(x)\geq x$

$f'(x) \geq 0$

etc. So another example function might be $f(p) = \sqrt{p}$.

If there's any additional information that would be useful, or if the problem as specified isn't clear, I'd be more than happy to edit the question as needed! Thank you!!

  • $\begingroup$ Have you looked at documentation of ForAll and Resolve? $\endgroup$ – kirma Sep 18 '18 at 17:55
  • $\begingroup$ @kirma I tried ForAll at one point, but it seemed to just kind of plug in the f functions into the "condition" equation and keep the rest of the conditions the same (specifically, I tried ForAll[{a, h, l}, condition[f, a, h, b, l] >= 0 && 0 <= a <= 1 && 0 <= l < h <= 1] so perhaps part of the issue is my incorrect specification of the problem). I hadn't tried using Resolve, but it's been running for the past 10 minutes now with no signs of progress (though again, I'm not entirely sure how long it should take). Thank you!! $\endgroup$ – AndrewC Sep 18 '18 at 18:10

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