Math Problem

[jimerax]

Challenging for you guys.. prove this.

if x^2+y^2+z^2=1 and x>0,y>0,z>0,

then yz/x+zx/y+xy/z>=3^(1/2)

[QreepyBORIS]

RE: Math Problem

y helo mr homework

[GuidoHunter]

That function is just a unit sphere with the constraints limiting it to the all-positive octant, right?

Even with that information I have no idea where to start. I've never been able to prove things with ingenuity.

--Guido

http://andy.mikee385.com

[jimerax]

That function is just a unit sphere with the constraints limiting it to the all-positive octant, right?

Yea right, but I think this fact doesn't help a lot for solution.

[QreepyBORIS]

Does somebody want to give me an example when it equals MORE than the square root of 3? -_-

Because it is painfully clear that it can equal the square root of three. :P

[GuidoHunter]

I'm just glad I could recognize that much. =)

Anyway, (1/√3, 1/√3, 1/√3) seems to be a minimum from plugging that into the equation. From this point, the only ways to find other points is to lower the value for one point and raise the value of the other two, vice versa, or lower one, raise another, and keep one the same.

So, raising one variable and lowering the other two to, say, 1/2, you get: (1/2, 1/2, √2/2). Plug that in and get 1.7677... > √3

Increasing two variables to, say, 1/√3+0.01, the point becomes: (1/√3+0.01, 1/√3+0.01, 0.556811748) and the result is something a little greater than √3.

Keep one at 1/√3, raise the other to an arbitrary value, then calculate the third and you get a point like (2/3, √2/3, 1/√3). Put that into the equation and get 1.76907.

So, with every possible point on that sphere the value is greater than or equal to √3.

That's as close to a proof as I can get.

--Guido

http://andy.mikee385.com

[Kilgamayan]

I recommend taking partial derivatives, setting them equal to zero, and finding maxima and minima.

I'd attempt it but I'm really lazy right now.

[GuidoHunter]

That was actually my first though, but I didn't want to bust out my CalIII notes. Plus, finding the maxima and minima is just intuitive.

--Guido

http://andy.mikee385.com

[Kilgamayan]

You need Calc III notes for partial derivatives? All you have to do is treat all other variables as constants.

And I hope you don't need Calc III notes for solving a system of three equations and three unknowns.

[GuidoHunter]

Nonono, I know how to take partial derivatives, but I can't remember what to do after that. I actually wrote it all down I could probably figure out what to do, but I didn't like Cal III, so it didn't stick too well.

--Guido

http://andy.mikee385.com

[bigsleytheoaf]

woah

I don't know why you guys are going on about calculus and whatnot, all you need to solve this is the AM-GM inequality:

AM-GM states:

If you have n numbers x1, x2, ... , xn then their arithmetic mean is greater than or equal to their geometric mean. In other terms:

(x1 + x2 + ... + xn) / n >= sqrt(x1 * x2 * ... * xn)

We only need to apply this twice to get our answer.

First, use AM-GM w/r/t the expression x^2 + y^2 + z^2:

(x^2 + y^2 + z^2)/3 >= sqrt(x^2 * y^2 * z^2) = xyz (since x, y, z all positive)

By substitution:

1/3 <= xyz

And therefore:

sqrt(1/3) <= sqrt(xyz) (i)

We'll use (i) after we apply the AM-GM to: xy/z + zy/x + xz/y as follows:

(xy/z + xz/y + zy/x)/3 >= sqrt(xy/z * zx/y * zy/x) = sqrt(x^2 * y^2 * z^2 / xyz)

->

xy/z + xz/y + zy/x >= 3 * sqrt(xyz) >= 3 * sqrt(1/3) = 3 / sqrt(3) = (3 * sqrt(3)) / 3 = sqrt(3)

Therefore, xy/z + xz/y + zy/x >= sqrt(3). Q.E.D.

(and no calculus in sight).

This is just a quick sketch and some details might be wrong, but I think that this is the general idea.

---

For more information on the AM-GM inequality:

http://mathworld.wolfram.com/ArithmeticMean.html

They use this inequality alot in math contests such as the Putnam, the AMC/AIME/USAMO/IMO track, etc. It's not terribly hard to prove either

-Bigsley

[blahblah18]

RE: woah

yeah... and awww I just got such fond childhood memories of AIME and USAMO... stupid me never getting to go to IMO boot camp, although my friend made it in IPO (the physics version) and got best female world I think.
Anyways, what I miss more then any of those easy contests is ARML ... anyone ever do that? Weekends at Penn State, so fun.

[Linkisdoomed]

RE: woah

Woa. What class did you all find this one in. Im in algebra. KTHNKSBAI I FAILD'. But really, I know this is not any form of an equation that I have seen before. *adjusts glasses for further inspection*

[jimerax]

Re: woah

I don't know why you guys are going on about calculus and whatnot, all you need to solve this is the AM-GM inequality:

AM-GM states:

If you have n numbers x1, x2, ... , xn then their arithmetic mean is greater than or equal to their geometric mean. In other terms:

(x1 + x2 + ... + xn) / n >= sqrt(x1 * x2 * ... * xn)

We only need to apply this twice to get our answer.

First, use AM-GM w/r/t the expression x^2 + y^2 + z^2:

(x^2 + y^2 + z^2)/3 >= sqrt(x^2 * y^2 * z^2) = xyz (since x, y, z all positive)

By substitution:

1/3 <= xyz

And therefore:

sqrt(1/3) <= sqrt(xyz) (i)

We'll use (i) after we apply the AM-GM to: xy/z + zy/x + xz/y as follows:

(xy/z + xz/y + zy/x)/3 >= sqrt(xy/z * zx/y * zy/x) = sqrt(x^2 * y^2 * z^2 / xyz)

->

xy/z + xz/y + zy/x >= 3 * sqrt(xyz) >= 3 * sqrt(1/3) = 3 / sqrt(3) = (3 * sqrt(3)) / 3 = sqrt(3)

Therefore, xy/z + xz/y + zy/x >= sqrt(3). Q.E.D.

I thought it is proved at the first time I saw it.. but it's wrong.

AM-GM inequality means
(x1 + x2 + ... + xn) / n >= (x1 * x2 * ... * xn)^(1/n)
not sqrt [^(1/2)].

In this case n=3, therefore
(x^2 + y^2 + z^2) / 3 >= (x * y * z)^(2/3)
is right.

AM-GM inequality is useful for solution of this problem though.

[bigsleytheoaf]

oh yeah

*shrug*

I think I actually did this problem once. I don't quite remember how though...



-Bigsley