# [✓] Simple inequality, MMA says there is an instance, can someone find one?

GROUPS:
 If r is real and c is complex, Mathematica says FullSimplify[Sqrt[1/(r+c)]Sqrt[(r+c)/(1+c)]] is not necessarily equal to Sqrt[1/(1+c)]. I cannot find an instance where it is not equal, and I cannot prove that Mathematica is wrong. I can't seem to use FindInstance to find such an instance. It either assumes both are complex, or, specifying Reals, that both are reals. Clear[r,c]; $Assumptions = {r \[Element] Reals}; FullSimplify[Sqrt[1/(r + c)] Sqrt[(r + c)/(1 + c)] - Sqrt[1/(1 + c)]] FindInstance[Sqrt[1/(r + c)] Sqrt[(r + c)/(1 + c)] != Sqrt[1/(1 + c)], {r, c}]  Answer 22 days ago 4 Replies  Try {r -> -2, c -> 1}  Answer 22 days ago  @Paul Reiser, Welcome to Wolfram Community! Please make sure you know the rules: https://wolfr.am/READ-1STThe rules explain how to format your code properly. If you do not format code, it may become corrupted and useless to other members. Please EDIT your posts and make sure code blocks start on a new paragraph and look framed and colored like this. int = Integrate[1/(x^3 - 1), x]; Map[Framed, int, Infinity]  Answer 22 days ago  Thank you for having patience with my silly question. In my problem, the real "r" was in fact Cosh[x] with x real, which is always >=1 which I neglected to consider. I was asleep at the wheel, and your example woke me up. Thank you. MMA still will not simplify, so now my problem is to find an instance which justifies it. Clear[c, Coshl];$Assumptions = {Coshl >= 1, c + Coshl != 0, 1 + c != 0}; FullSimplify[ Sqrt[1/(c + Coshl)] Sqrt[(c + Coshl)/(1 + c)] - Sqrt[1/(1 + c)]] 
 Since square root function does not automatically simplify Sqrt[x y}] to Sqrt[x] Sqrt[y],we may use the PowerExpand code: \$Assumptions = c \[Element] Complexes; PowerExpand[Sqrt[1/(c + Coshl)] Sqrt[(c + Coshl)/(1 + c)]] // Simplify PowerExpand[Sqrt[1/(c + Coshl)] Sqrt[(c + Coshl)/(1 + c)]-Sqrt[1/(1+c)]]