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Take the following set of inequalities:

Reduce[
  {
    ys < Power[xs^a + ys^a - x^a, 1/a],
    0 < xs < 10000,
    0 < ys < 12000,
    0 < x <10000,
    xs < x,
    0 < a < 1
  },
  {xs, ys, x, a},
  Reals
]

When I run it, I get the following error:

This system cannot be solved with the methods available to Reduce.

But why? How can I make it solvable via Reduce?

I read this other thread: Can Reduce really not solve for x here?, and I understood that if you provide a specific domain, as I did, the system should become solvable even if it involves transcendental functions.

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  • 2
    $\begingroup$ How is Pow defined? $\endgroup$ – DiSp0sablE_H3r0 May 20 at 10:50
  • 2
    $\begingroup$ I think you overstate the ability of Reduce/Solve over a compact domain. This blog post indicates that for a single-variable holomorphic function, an equation can be solved over a compact domain. The situation for multivariate functions or ones that are not holomorphic is less clear to me. My experience is that some of them cannot be solved, so maybe not yours. The function z^a, with a a real variable and z a real or complex variable, is a common obstruction. $\endgroup$ – Michael E2 May 20 at 11:51
  • $\begingroup$ @DiSp0sablE_H3r0 Whoops, I thought that that is the standard Power function. Yet, even after replacing Pow with Power, I am getting the same error. $\endgroup$ – Paul Razvan Berg May 20 at 16:34
  • $\begingroup$ Thanks for the heads-up, @MichaelE2. $\endgroup$ – Paul Razvan Berg May 20 at 16:44
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    $\begingroup$ Something can be done:NMinimize[{a, ys < Power[xs^a + ys^a - x^a, 1/a], 0 < xs < 10000, 0 < ys < 12000, 0 < x < 10000, xs < x, 0 < a < 1}, {xs, ys, x, a}] produces {0.00912988, {xs -> 2426.64, ys -> 0.022711, x -> 5341.51, a -> 0.00912988}} and NMaximize[...] produces 0.32545, {xs -> 0.5, ys -> 0.5, x -> 0.5, a -> 0.32545}}. Unfortunately, then FindInstance[ ys < Power[xs^a + ys^a - x^a, 1/a] && 0 < xs && 0 < ys < 12000 && 0 < x < 10000 && xs < x /. a -> 1/5, {xs, ys, x}, Reals] is running for ages. $\endgroup$ – user64494 May 20 at 19:08
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If you change to non-powered variables {xs1,ys1,x1}, you get solutions for at least rational a of the form a = 1/( 2 b) with b being positive integers.

First get conditions for changed variables and then test your inequation.

{a = 1/Pi, 
red = List @@ 
Reduce[{xs^a == xs1, ys^a == ys1, x^a == x1, 0 < xs <  10000, 
  0 < ys < 12000, 0 < x < 10000, xs < x}, {xs, ys, x}, Reals] // 
 PowerExpand}

(*   {1/\[Pi], {0 < ys1 < 2^(5/\[Pi]) 3^(1/\[Pi]) 5^(3/\[Pi]), 
0 < x1 < 10^(4/\[Pi]), 0 < xs1 < x1, xs == xs1^\[Pi], 
ys == ys1^\[Pi], x == x1^\[Pi]}}   *)

Reduce[ys1^(1/a) < Power[xs1 + ys1 - x1, 1/a] && 
   And @@ red[[1 ;; 3]], {xs1, ys1, x1}, Reals]

(*   Reduce::nsmet: This system cannot be solved with the methods available to Reduce. >>   *)

With a = 3/10 you get "False". And so on. Let make a test function.

test := {a = Rationalize[RandomReal[{.06, .94}], 1/20], 
  red = List @@ 
Reduce[{xs^a == xs1, ys^a == ys1, x^a == x1, 0 < xs < 10000, 
  0 < ys < 12000, 0 < x < 10000, xs < x}, {xs, ys, x}, Reals]; 
Reduce[ys1^(1/a) < Power[xs1 + ys1 - x1, 1/a] && 
And @@ red[[1 ;; 3]], {xs1, ys1, x1}, Reals]}

Table[test, {20}] // TableForm

Result ( not shown here) indicates there are only solutions for a = 1/ (2 b) with b beeing positive integers. I leave it to you to prove this.

Graphics says the same.

Clear[a]; Manipulate[
RegionPlot3D[
ys < Power[xs^a + ys^a - x^a, 1/a] && xs < x, {xs, 0, 10000}, {ys, 
0, 12000}, {x, 0, 10000}] // 
Quiet, {a, {1/2, 1/E, 1/Pi, 1/3, 2/3, 1/4, 1/6, 1/8, 1/10, 1/12, 
1/14, 1/16, 1/18}}]
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