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I have an inequality as follows

2^(1/2 (1 + 1/n)) > 0 && 
 t <= (2^(1/2 (-1 - 1/n) + n/2) n^(1 + 1/(2 n)) \[Pi]^((1/2)/n))/E

I want to simplify the 2^(1/2 (1 + 1/n)) > 0 to True using assumption that n > 0.

However, if I do the following,

2^(1/2 (1 + 1/n)) > 0 && 
  t <= (2^(1/2 (-1 - 1/n) + n/2) n^(1 + 1/(2 n)) \[Pi]^((1/2)/n))/E //
  FullSimplify[#, n > 0] &

I end up with

2^(1/2 (1 + 1/n)) E t <= 2^(n/2) n^(1 + 1/(2 n)) \[Pi]^((1/2)/n)

But I want to keep the t on one side of inequality. How can I do that.

Note the example is a bit simplified. I have a much complicated expression which I get from Reduce which I want to simplify, while keep t isolated on one side of inequalities.

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  • $\begingroup$ Let us observe that 2^(1/2 (1 + 1/n)) is positive at any n. Therefore, you can safely replace your inequality by t <= (2^(1/2 (-1 - 1/n) + n/2) n^(1 + 1/(2 n)) \[Pi]^((1/2)/n))/E &&n > 0, and work with this one. You did not describe, though, what are you expecting to get from it? Limitations on what variable do you want to obtain? $\endgroup$ Oct 14 at 11:30
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Clear["Global`*"]

expr = 2^(1/2 (1 + 1/n)) > 0 && 
   t <= (2^(1/2 (-1 - 1/n) + n/2) n^(1 + 1/(2 n)) π^((1/2)/n))/E;

expr2 = ReplacePart[expr, 1 -> Simplify[expr[[1]], n > 0]]

(* t <= (2^(1/2 (-1 - 1/n) + n/2) n^(1 + 1/(2 n)) π^((1/2)/n))/E *)
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  • $\begingroup$ Why do we need to clear "*"? $\endgroup$
    – ablmf
    Oct 14 at 4:59
  • $\begingroup$ It clears the Global` namespace to ensure that there are no old definitions laying around (i.e., for n or t) that might interfere with the subsequent code. $\endgroup$
    – Bob Hanlon
    Oct 14 at 5:03
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Or

expr = 2^(1/2 (1 + 1/n)) > 0 && 
       t <= (2^(1/2 (-1 - 1/n) + n/2) n^(1 + 1/(2 n)) \[Pi]^((1/2)/n))/E;

expr // Refine[#, Assumptions -> n > 0] &

(*   t <= (2^(1/2 (-1 - 1/n) + n/2) n^(1 + 1/(2 n)) \[Pi]^(1/(2 n)))/E   *)

And with definite n

expr // Refine[#, n == 2] &

(*   t <= (2 Sqrt[2] \[Pi]^(1/4))/E   *)
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