# No real solution for 3 coupled equations, but then how "close" can one get?

I have three equations involving $$x$$, $$y$$, and $$z$$ that I would like to find a unique solution to, in the Reals.

NSolve does not find anything:

NSolve[{Sqrt[1 / (x^4)] == Sqrt[1 /((y^3) z )],
Sqrt[1 / (x^4)] == Sqrt[(1/x^4 + 1 /( y z^3))],
Sqrt[1 /( (y^3) z )] == Sqrt[(1 /x^4 + 1 /(y z^3))]}, {x, y,
z}, Reals]


which seems confirmed by a CountourPlot3D which shows no intersection:

ContourPlot3D[{Sqrt[1 / (x^4)] == Sqrt[1 /((y^3) z )],
Sqrt[1 / (x^4)] == Sqrt[(1/x^4 + 1 /( y z^3))],
Sqrt[1 /( (y^3) z )] == Sqrt[(1 /x^4 + 1 /(y z^3))]}, {x, 0,
100}, {y, 0, 100}, {z, 0, 100}, PlotLegends -> "Expressions"]


Question: is there a way to find "approximate" solutions, specifying an accuracy range within which a solution could be? I.e. a particular (x,y,z) combo obviously doesn't give 0 to one of my equations, but maybe gives 0.1 and that'd be good enough...

Clear["Global*"]


Reformulate as a minimization problem

expr = Subtract @@@ {Sqrt[1/(x^4)] == Sqrt[1/((y^3) z)],
Sqrt[1/(x^4)] == Sqrt[(1/x^4 + 1/(y z^3))],
Sqrt[1/((y^3) z)] == Sqrt[(1/x^4 + 1/(y z^3))]}

(* {Sqrt[1/x^4] - Sqrt[1/(y^3 z)],
Sqrt[1/x^4] - Sqrt[1/x^4 + 1/(y z^3)], -Sqrt[1/x^4 + 1/(y z^3)] + Sqrt[1/(
y^3 z)]} *)


Minimizing the sum of the squares

{min, arg} =
NMinimize[Join[{Total[expr^2]}, Thread[0 <= {x, y, z} <= 100]], {x, y, z},
Method -> "DifferentialEvolution",
WorkingPrecision -> 17]

(* {1.2736005375588001*10^-9, {x -> 44.219798564468166, y -> 32.928582883862153,
z -> 99.999999784859630}} *)

expr /. arg

(* {-0.0000178183282646516, -0.0000288761531935047, -0.0000110578249288531} *)

• Note that increasing the allowed range of x,y,z seems to reliably yield a solution of the form $x \approx 0.4 z, y \approx 0.3 z$, and $z$ maxed out within the allowed range. Doing so also decreases the value of min`. There's probably a way to prove this analytically, but I don't see it immediately. Nov 23, 2021 at 19:32