Theorem about two real numbers 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.
 A: 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.
A: We have 
$$(a+b)^2=(a-b)^2+4ab=(a-b)^2+4k.$$
To minimize $a+b$, we minimize $(a+b)^2$. To do this, we minimize $(a-b)^2$, by setting $a=b$.
A: Hint: $ab = k \Rightarrow b = k/a$
Then to minimize $S = a + b = a + k/a$, take derivatives. 
A: Let us rephrase the problem: find the minimum of $\,x+y\,$ when $$x,y\in\mathbb{R}^+\,\,,\,\,and\,\,xy=k=constant$$
We already get that $\,\displaystyle{y=\frac{k}{x}}\,$ , so we need the minimum of the function$$f(x)=x+y=x+\frac{k}{x}\Longrightarrow f'(x)=1-\frac{k}{x^2}=0\Longleftrightarrow x=\pm|k|$$
Well, check with the derivative's changes of signs or with the second derivative test which one is what.
