7
$\begingroup$

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.

4 Answers 4

13

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 $a=b$.

  • 0
    @ Marvis: Thanks for that2012-06-07
4

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.

1

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

  • 0
    Just as long as I don't get more than five downvotes for each upvote, I'll be in the black :)2012-06-07
1

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.