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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,490

HI;

With

Prove:

2)

With

Prove:

*Last edited by bobbym (2009-09-15 13:43:27)*

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

Wow, they sure look nasty! (At first sight, at any rate.)

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,490

Hi Jane;

Their supposed to be super tough but...

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

Well, I can at least start by proving that .

Quite easy, as it turns out.

In fact, as you can see, the inequality is strict.

So far so good.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,490

Hi Jane;

Pretty!

Here is my solution for the rest:

Did sub c = 1 - a - b

Hold b constant

So a is at maximum when a = (1 - b) / 2. If we now substitue the maximum value for a and the inequality still holds then it holds for lesser values of a. Substituting into

yields with simplification

Do the switch cause of multiplication by - 27

The above is obviously true, so

Hope yours is shorter!

*Last edited by bobbym (2009-09-16 02:15:07)*

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

I always thought using calculus to tackle these kinds of inequalities is a bit of a brute-force tactic. Much like using the JordanHölder theorem in group theory to prove the fundamental theorem of arithmetic, if you know what I mean.

But of course, if calculus works, then there is no reason not to use it, if you know how. Nice work!

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,490

Hi Jane;

I give the proof (mine) a two.

The author of the book that these come from has said that the olympiad and putnam committees who judge the answers agree with you. He feels that all these types of problems are especially vulnerable to maxima - minima. I agree but prefer not to use it, can you come up with some other way?

It is funny though that mathematicians who supposedly favor general methods ( the teakettle principle ) insist on using a different method for each of these problems. As I have said I also prefer other methods.

*Last edited by bobbym (2009-09-17 08:06:02)*

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,490

Hi;

Another toughie:

x>0 , y>0, z>0

Prove:

*Last edited by bobbym (2009-10-16 18:23:15)*

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,490

Hi Jane;

That was like a slap in the face with a wet towel. Good job!

You probably looked this over already but in case you haven't:

https://nrich.maths.org/discus/messages … -69859.pdf

Here are some interesting comments about it from the AOPS forum.

According to Andre Rzym, it's a useful tool, and a good way to intimidate people who aren't familiar with it!

A common bruteforce technique with inequalities is to clear denominators, multiply everything out, and apply Muirhead's or Schur's. However, it is worth noting that any inequality that can be proved directly with Muirhead can also be proved using the Arithmetic Mean-Geometric Mean inequality. In fact, IMO gold medalist Thomas Mildorf says it is unwise to use Muirhead in an Olympiad solution; one should use an application of AM-GM instead. Thus, it is suggested that Muirhead be used only to verify that an inequality can be proved with AM-GM before demonstrating the full AM-GM proof.

Using a lot of that bruteforce myself I am forced to disagree with their comments. I like it a lot!

*Last edited by bobbym (2009-10-17 12:50:00)*

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**Identity****Member**- Registered: 2007-04-18
- Posts: 934

Wow, Muirhead's inequality looks quite powerful, thanks for introducing me to it

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,490

Hi identity;

Jane did the introductions, don't tell Jane, but due to my laziness I wasn't familiar with it.

*Last edited by bobbym (2009-10-16 23:53:07)*

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**Identity****Member**- Registered: 2007-04-18
- Posts: 934

Haha, unwise to use Muirhead in an IMO solution, because it is not elegant enough?

How about proving Muirhead's yourself in the exam and acting as if you didn't know it existed! That would seem like real genius.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,490

Hi identity;

I agree, I think their comments are weird. I am not being impolite, saying weird, I just don't understand. How many times do you have the luxury of choosing methods by some criterion other than what works?

*Last edited by bobbym (2009-10-17 12:47:21)*

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,490

Hi;

Prove:

*Last edited by bobbym (2009-10-22 11:25:08)*

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

I think you have the inequality sign the wrong way round.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,490

Hi Jane;

You solved it. It's supposed to be false. Good answer.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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