# 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.

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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 $a=b$.

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I like this answer better :) – The Chaz 2.0 Jun 7 '12 at 22:59
@TheChaz: The two answers complement each other. – André Nicolas Jun 7 '12 at 23:01
Yes, this one's more elegant. – DonAntonio Jun 7 '12 at 23:04
@meg_1997: Since $a+b$ must be positive, we minimize $a+b$ precisely if we minimize $(a+b)^2$. But $(a+b)^2=(a-b)^2+4k$. Since $k$ is fixed, the right-hand side is minimized by making $(a-b)^2$ as small as possible. But $(a-b)^2\ge 0$ always, so it is a minimum when it is $0$, that is, when $a=b$. Since $ab=k^2$, putting $a=b$ we get $b^2=a^2=k$, so $b=a=\sqrt{k}$. – André Nicolas Jun 7 '12 at 23:13
@meg_1997 What André has written is nothing but the proof of AM-GM for two real numbers $a$ and $b$ and argued out why equality in AM-GM occurs when $a=b$. To rewrite André identity in AM-GM form, divide André equation by $4$ to get that $$\left( \dfrac{a+b}2\right) = \sqrt{ab + \left(\dfrac{a-b}2 \right)^2}$$ and since $\left(\dfrac{a-b}2 \right)^2 \geq 0$, we get that $$\left( \dfrac{a+b}2\right) \geq \sqrt{ab}$$ with equality holding when $\left(\dfrac{a-b}2 \right)^2 = 0$ i.e. when $a=b$. – user17762 Jun 7 '12 at 23:47

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.

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Hint: $ab = k \Rightarrow b = k/a$
Then to minimize $S = a + b = a + k/a$, take derivatives.

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Just as long as I don't get more than five downvotes for each upvote, I'll be in the black :) – The Chaz 2.0 Jun 7 '12 at 23:51

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.

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