How can I solve this inequality using Mathematica?

Sqrt[B^2 + J^2] Cosh[(2 J)/T ] < J Sinh[  (2 Sqrt[B^2 + J^2])/T]

where T, J, and B are reals and they are >0.

  • 2
    $\begingroup$ I doubt that something that complicated can be solved analytically at all. Reduce[ { Sqrt[B^2 + J^2] Cosh[(2 J)/T] < J Sinh[(2 Sqrt[B^2 + J^2])/T], {B, J, T} > 0}, {B, J, T} ] quickly returns saying that "This system cannot be solved with the methods available to Reduce". $\endgroup$ – MarcoB Jul 7 '20 at 18:42

You can divide both sides by T. This reduces the number of variables to 2: $$ j=\frac{J}{T},\quad b=\frac{B}{T}.$$

To visualize inequality, one can use ContourPlot. The inequality is fulfilled in the green area.

ContourPlot[Sqrt[b^2 + j^2] Cosh[2 j] - j Sinh[2 Sqrt[b^2 + j^2]],{j,0,1},{b,0,1},
ColorFunction -> (If[# > 0., RGBColor[0,0.5+#/2, 0], GrayLevel[(1 - #)/3]] &)]

enter image description here


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