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mohd

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show that

luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

Re: show that

try making b and c dependant on a. like:

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mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: show that

Luca has the right idea, although it might be easier if you say that c = a+h.

You can also say without loss of generality that w>h>0, if that helps at all.

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Mohd

Guest

Re: show that

i thought about what you said ... but i couldnt solve it yet

Stanley_Marsh

Member

Registered: 2006-12-13

Posts: 345

Re: show that

Ahh, Haven't worked it out yet , but it's so close

Now it's left to compare

WOW , What happen to my post , Something wrong with the server?

Last edited by Stanley_Marsh (2007-04-14 14:33:27)

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Numbers are the essence of the Universe

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: show that

Great work! But I don’t think you can proceed further, unfortunately.

I’ve been testing with various values of a. b. c on an Excel spreadsheet. It is true that A ≥ P and A ≥ Q (with the values I’ve tried), but I’ve found that for some values your Q > P and for other values P > Q.

I ran into a similar brick wall yesterday. I found that A ≥ some expression R, and I thought maybe R ≥ P as well, so I could prove A ≥ P via that route. Indeed, I tried a lot of values on my Excel spreadsheet and found that R ≥ P for those values. Unfortunately, I suddenly discovered a set of values for which R < P – so all my hard work came to nothing.

Stanley_Marsh

Member

Registered: 2006-12-13

Posts: 345

Re: show that

To do this question , you need to go through series of transformation and apply different inequality. This kind of question appeared everytime in my test in China, but it's my weakness , lol. And I am sure this one is on the Olympiad level.

Last edited by Stanley_Marsh (2007-04-15 05:14:10)

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Numbers are the essence of the Universe

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: show that

All right, I think we have no choice but to use calculus. Let

For maximum or minimum E, all the first-order partial derivatives must be 0.

We don’t need to work out all the partial derivatives because, see? E is symmetrical in a, b, c. Therefore whatever critical values we find, we can be sure that it will have a = b = c. So substitute a = b = c into the first partial-derivative equation above.

Solving this gives a = 0, ±1, and two complex roots. We are given that a, b, c > 0. So a = b = c = 1, which gives E = 0. It remains to show that (1,1,1) does give a minimum point. Since there are no other critical values in the octant a, b, c > 0, this would make it a global minimum in this octant – i.e. E ≥ 0 for a, b, c > 0.

So how do we show that this is a minimum point? Any ideas? (Someone at the PlanetMath site said I could use the Hessian but I’m not sure what this is.)

Last edited by JaneFairfax (2007-04-20 10:39:44)

Zhylliolom

Real Member

Registered: 2005-09-05

Posts: 412

Re: show that

Hessian is like the Jacobian (without further qualification they are determinants, we can also consider the Hessian or Jacobian matrix, which are useful as the matrix of certain transformations, such as change of coordinates), but its elements are second order derivatives. The Hessian is defined as

Suppose a = (a[sub]1[/sub], ..., a[sub]n[/sub]) is a critical point of f. Then the following hold:

In more advanced terms, if the Hessian matrix is positive definite at a, then a is a local minimum for f, if the Hessian matrix is negative definite at a, then a is a local maximum for f, and if the Hessian matrix has both positive and negative eigenvalues at a, then a is a saddle point of f. This is basically the second derivative test for functions of more than one variable. Note that this second derivative test may sometimes be inconclusive.

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: show that

Thanks.

So

and thanks to the symmetry of E the other values will be the same. So I get the Hessian at (1,1,1) as

This has a positive determinant and also ∂[sup]2[/sup]E/∂a[sup]2[/sup] > 0 at (1,1,1).

So, is that it? Have I shown that E has a minimum at (1,1,1)? Has the problem, after nearly one long week, finally been solved?

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: show that

In fact, the Hessian at (−1,−1,−1) is the same as at (1,1,1), so (−1,−1,−1) also gives a minimum point, and E(1,1,1) = E(−1,−1,−1) = 0. Apllying the Hessian test for (0,0,0) shows that (0,0,0) gives a saddle point. So the minimum points are global minima. Therefore

Last edited by JaneFairfax (2007-04-20 20:47:37)

luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

Re: show that

corrrection?

Last edited by luca-deltodesco (2007-04-20 17:44:06)

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The Beginning Of All Things To End.
The End Of All Things To Come.

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: show that

Zhylliolom

Real Member

Registered: 2005-09-05

Posts: 412

Re: show that

Haha, I looked at that kind of funny when you wrote it, and wondered if I had helped at all when you arrived at a trivially obvious conclusion. Good work, team.

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: show that

BUMP!

I remember this problem. It gave me a lot of headache all that time (nearly two years) ago. But now I think I can solve it without the hassle of calculus.

Let

. So we want to prove this equivalent inequality

However, note that

(Cauchy–Schwarz)

.

We will therefore prove the following stronger inequality (which implies the one above):

i.e.

Proof:

Expanding the LHS gives

This can be rewritten as

Obviously

. Thus we are done if we can show that

.

Well, then.

Treating this as a quadratic in

, the discriminant is

This proves that

. QED!!

Last edited by JaneFairfax (2009-01-08 14:53:15)