1
$\begingroup$

Am I missing something here? This does not resolve to True.

Clear[x,y]

$Assumptions={x>0,y>0}

FullSimplify[(Sqrt[x] + Sqrt[y]) == Sqrt[x + y + 2 Sqrt[x y]]]
$\endgroup$
2
  • $\begingroup$ Simplify[Reduce[{(Sqrt[x] + Sqrt[y]) == Sqrt[x + y + 2 Sqrt[x y]], x > 0 && y > 0}], {x > 0, y > 0}] $\endgroup$
    – cvgmt
    Apr 8, 2022 at 8:17
  • $\begingroup$ I think this is limitation in Simplify because Reduce[(Sqrt[x] + Sqrt[y]) - Sqrt[x + y + 2 Sqrt[x y]] == 0, Reals] gives y >= 0 && x >= 0, therefore Simplify with this assumption should give True but it does not. $\endgroup$
    – Nasser
    Apr 8, 2022 at 8:23

1 Answer 1

1
$\begingroup$
Resolve[ForAll[{x, y}, 
  Sqrt[x] + Sqrt[y] == Sqrt[x + y + 2 Sqrt[x y]]], PositiveReals]

True

$\endgroup$

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