I defined a long expression

exp = (-1+ψ) ψ (-1+τ+ψ-2 τ ψ)^2 (1-2 τ)^2+(-1+τ+ψ-2 τ ψ)^2 (-1+τ+ψ)^2 (1-ψ+τ (-1+2 ψ))-(τ-ψ)^2 (τ+ψ-2 τ ψ) (2+τ^2 (1-2 ψ)^2+(-4+ψ) ψ-2 τ (-2+ψ) (-1+2 ψ))

and with some specified conditions, Mathematic tells me the expression is strictly positive

  exp > 0
 ,1 > τ > 1/2 && 1 > ψ > 1/2 && vh > vl > 0 && γ > 0 && τ < ψ
(* True *)

I want to get a proof for it. Is there any way to let Mathematica help me find the "arranged" version of this expression to show the sign?

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    $\begingroup$ Does Reduce[exp > 0, {\[Psi], \[Tau]}] help you? (You have several irrelevant assumptions in your Simplify command, btw.) $\endgroup$ – Michael E2 Jul 23 at 15:52
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    $\begingroup$ @MichaelE2 Thanks for editing the question. I want to organize the expression in a way that we can "directly see" why it is positive, given the parameter conditions. For exameple, for expression $y=x^2 - 2x +2$, we can write it as $y=(x-1)^2 + 1 >0$. I tried Reduce[] but it does not do this job. $\endgroup$ – David Xiaoyu Xu Jul 23 at 17:19

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