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I'm trying to solve an inequality in terms of one variable under some conditions for parameter. Here is the code.

Assuming[lm > 1 && 0 < u < 1 && rmax > 0 && iD > 0 && rmax > iD && iE > iD, FullSimplify@Reduce[(iE + rmax + lm rmax (-1 + u) - iD lm u)/(1 - lm) - iD > 0, lm]]

The code instructs to solve in terms of lm but the outcome is in terms if iE, which is as follows.

(iD - rmax) (1 + lm (-1 + u)) > iE

Do you know why?

**EDIT: user64494's answer suggested below does not work either. To see this, consider the following inequality:

Assuming[(iD - rmax) (1 + lm (-1 + u)) > iE && lm > 1 + lm u && iE + rmax u < rmax + iD u && iE lm + rmax + lm rmax u < lm (rmax + iD u) && lm > 1 && 0 < u && u < 1 && rmax > 0 && iD > 0 && rmax > iD && rmax > iE && iE > iD, FullSimplify@Reduce[rmax/2 + iE > rmax + iE/(1 + lm (-1 + u)), rmax]]

whose outcome is

lm (2 iE - rmax) (-1 + u) < rmax

which is not exactly in terms of rmax. Now, following user64494's approach:

Reduce[(iD - rmax) (1 + lm (-1 + u)) > iE && lm > 1 + lm u && iE + rmax u < rmax + iD u && iE lm + rmax + lm rmax u < lm (rmax + iD u) && lm > 1 && 0 < u && u < 1 && rmax > 0 && iD > 0 && rmax > iD && rmax > iE && iE > iD && rmax/2 + iE > rmax + iE/(1 + lm (-1 + u)), rmax]

whose outcome is

iE > 0 && 0 < iD < iE && 0 < u < iE/(-iD + 2 iE) && lm > -(1/(-1 + u)) && (-iE lm + iD lm u)/(1 - lm + lm u) < rmax < (-2 iE lm + 2 iE lm u)/(1 - lm + lm u)

from which I cannot immediately tell which one is the solution for the inequality I'm looking for.

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  • $\begingroup$ Reduce[(iE + rmax + lm rmax (-1 + u) - iD lm u)/(1 - lm) - iD > 0, lm, Reals] $\endgroup$
    – csn899
    Commented Nov 6, 2023 at 23:52
  • $\begingroup$ @csn899, I get the same answer. $\endgroup$
    – ppp
    Commented Nov 6, 2023 at 23:55

1 Answer 1

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The following works in 13.3.1 on Windows 10.

Reduce[(iD - rmax) (1 + lm (-1 + u)) > iE && lm > 1 && 0 < u < 1 && 
rmax > 0 && iD > 0 && rmax > iD && iE > iD, lm]

rmax > 0 && 0 < iD < rmax && iE > iD && 0 < u < 1 && lm > (-iD + iE + rmax)/(-iD + rmax + iD u - rmax u)

Addition.@ppp: Don't understand what you mean by "complete answer". The result can be rewritten as Piecewise[{{lm > (-iD + iE + rmax)/(-iD + rmax + iD u - rmax u), lrmax > 0 && 0 < iD < rmax && iE > iD && 0 < u < 1}, {Null, True}}].

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  • $\begingroup$ Thanks, but it does not provide a complete answer. I added why in my post in the edit. Can you please take a look and help me once more? Thanks! $\endgroup$
    – ppp
    Commented Nov 8, 2023 at 1:06

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