# Simplify of -Sqrt[r] >= 0 where r is real

GROUPS:
 Am I missing something? Assuming r is real, the only r that can satisfy -Sqrt[r]>=0 is zero. \$Assumptions = {r \[Element] Reals}; FullSimplify[-Sqrt[r] >= 0] yields Sqrt[r] <= 0rather than r==0
 Michael Rogers 1 Vote Either Reduce[-Sqrt[r] >= 0] or Simplify[-Sqrt[r] >= 0, r \[Element] Reals, TransformationFunctions -> {Automatic, Reduce}] will produce r == 0 The transformations available to Simplify and FullSimplify are limited to some extent. See the documentation of FullSimplify for a fuller example of using Reduce as a transformation function.