TITLE: Theorem about two real numbers
QUESTION [7 upvotes]: My question is:
$a,b$ are two positive real numbers such that their product is constant,equal to $k$ say. Prove: the sum $a+b$ is minimum if and only if $a = b= \sqrt k$.
Can this be solved using $A.M.-\;G.M.$ inequality? If yes,then I would like to know it that way too.

REPLY [4 votes]: Since you asked specifically for a proof using the A.M.-G.M. inequality, I'll provide one, even though it's not as elegant or direct as the proof in André Nicolas's answer (which does not use the A.M.-G.M. inequality).
By the A.M.-G.M. inequality, $\sqrt{ab}\le\frac{a+b}2$,
i.e. $2\sqrt{ab}\le a+b$, with equality iff $a=b$.
Since the left-hand side of that inequality is fixed ($2\sqrt k$), we have that the right-hand side is equal to a fixed number it is not less than — i.e. is minimized — iff $a=b$, as sought.