Does there exist a constructive proof that the arithmetic mean equals the geometric mean if and only if all arguments are equal, without relying on excluded middle or dependent choice?

The arithmetic mean-geometric mean inequality is a cornerstone of classical analysis, and its equality condition, that AM equals GM precisely when all inputs are identical, is treated as trivial in classical mathematics. However, in constructive mathematics, where every existence claim must come with an explicit procedure and the law of excluded middle is not assumed, the equality condition becomes genuinely problematic. To constructively prove that equal means implies equal arguments, one needs to extract a witness pointing to which argument differs from the others whenever they are not all equal, and no such extraction procedure is currently known to be uniformly computable.