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 Post subject: Ineq Questions that End Up Proving a Very Nice IneqPosted: Fri, 6 Apr 2012 00:37:42 UTC
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Joined: Wed, 4 Apr 2012 03:51:40 UTC
Posts: 129
Location: Hockeytown aka Detroit
Information needed to know:

AM-GM, Cauchy-Schwartz, Schur's, Rearrangement, Jensen's, Muirhead's

Problems:
1.Show that for all positive :

Spoiler:
It's equivalent to which is just AM-GM.

2.Prove that for all positive :

I wasn't sure how to do how to do this one.~

3.Prove that for all positive :

Darn couldn't get this one either.

4.Let be nonnegative reals such that

Then prove:

Solution Sketch:

Spoiler:
Assume for the sake of contradiction that this is not the case.

Let where .

Similarly define , where .

Let be the unique positive integer satisfying

What we wish to show is that:

Since both the RHS and LHS are positive, this is equivalent to:

So we just need to show that:

It suffices to show:

Thus, we want to prove that:

Since the RHS of the above equation is greater than , the result follows iff:

As a result, it suffices to show that:

The above inequality is true iff:

Or:

But, by repeating the above procedure the exponent of the "explodes" since is a fixed yet arbitrary positive integer it must be passed eventually.

This is a contradiction and our lemma is true.

5. Let be positive real numbers. Prove that

It's supposed to work so that we prove 1-4 and then 5. Unfortunately I can't show 2 and 3.

Unrelated Inequality:

Suppose that the real numbers satisfy:

Determine the maximum of:

and when equality occurs

EDIT:Typo

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 Posted: Fri, 6 Apr 2012 01:18:55 UTC
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 13544
Location: Austin, TX
rdj5933mile5math64 wrote:
Information needed to know:

AM-GM, Cauchy-Schwartz, Schur's, Rearrangement, Jensen's, Muirhead's

Problems:
1.Show that for all positive :

Spoiler:
It's equivalent to which is just AM-GM.

2.Prove that for all positive :

I wasn't sure how to do how to do this one.~

3.Prove that for all positive :

Darn couldn't get this one either.

4.Let be nonnegative reals such that

Then prove:

Solution Sketch:

Spoiler:
Assume for the sake of contradiction that this is not the case.

Let where .

Similarly define , where .

Let be the unique positive integer satisfying

What we wish to show is that:

Since both the RHS and LHS are positive, this is equivalent to:

So we just need to show that:

It suffices to show:

Thus, we want to prove that:

Since the RHS of the above equation is greater than , the result follows iff:

As a result, it suffices to show that:

The above inequality is true iff:

Or:

But, by repeating the above procedure the exponent of the "explodes" since is a fixed yet arbitrary positive integer it must be passed eventually.

This is a contradiction and our lemma is true.

5. Let be positive real numbers. Prove that

It's supposed to work so that we prove 1-4 and then 5. Unfortunately I can't show 2 and 3.

Unrelated Inequality:

Suppose that the real numbers satisfy:

Determine the maximum of:

and when equality occurs

EDIT:Typo

I'm not sure "Olympiad level" is something that can be said for inequalities--AM-GM and almost all the others you listed are extremely standard. Perhaps you mean the application is somehow nonstandard or more difficult to see.

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 Posted: Fri, 6 Apr 2012 01:24:18 UTC
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Joined: Wed, 4 Apr 2012 03:51:40 UTC
Posts: 129
Location: Hockeytown aka Detroit

EDIT: what I mean by olympiad level is high school mathematical proof contests similar in difficulty to the usamo

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 Posted: Fri, 6 Apr 2012 04:34:20 UTC
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Joined: Mon, 29 Dec 2008 17:49:32 UTC
Posts: 6616
Location: On this day Taiwan becomes another Tiananmen under Dictator Ma.
rdj5933mile5math64 wrote:
2.Prove that for all positive :

I wasn't sure how to do how to do this one.~

I don't know either. What am I supposed to prove with ?

rdj5933mile5math64 wrote:
3.Prove that for all positive :

Darn couldn't get this one either.

Assuming means cyclic sum, not symmetric sum:
Spoiler:
Move everything to the LHS and clearing denominators, it suffices to prove

which is just an instance of the Muirhead's inequality.

For more about Muirhead's inequality, consult wikipedia/AoPS/... It is an inequality that trivialises a very large proportion of olympiad inequalities.

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 Posted: Fri, 6 Apr 2012 05:47:18 UTC
 Senior Member

Joined: Wed, 4 Apr 2012 03:51:40 UTC
Posts: 129
Location: Hockeytown aka Detroit
outermeasure wrote:
Assuming means cyclic sum, not symmetric sum:

That's correct I should have been more clear.

outermeasure wrote:
rdj5933mile5math64 wrote:
2.Prove that for all positive :

I wasn't sure how to do how to do this one.~

I don't know either. What am I supposed to prove with ?

ehehe sorry the correct inequality is below

outermeasure wrote:
rdj5933mile5math64 wrote:
3.Prove that for all positive :

Darn couldn't get this one either.

Spoiler:
Move everything to the LHS and clearing denominators, it suffices to prove

which is just an instance of the Muirhead's inequality.

For more about Muirhead's inequality, consult wikipedia/AoPS/... It is an inequality that trivialises a very large proportion of olympiad inequalities.

Awesome! Thank You!

And while I'm at it:
rdj5933mile5math64 wrote:
Unrelated Inequality:

Suppose that the real numbers satisfy:

Determine the maximum of:

and when equality occurs

Solution Sketch:
Spoiler:
Multiplying the second given inequality by and using the first inequality yields:

Multiplying the third given inequality by and using the first inequality yields:

So, we add the inequalities to get that:

Note that the maximum is reached when the inequalites turn into equalities so
.

From this we get that

So, by the first equation either which means that or .

Now, let be a positive integer such that . Observe that: and

Also note that . So, if , then by the first given inequality everything else is . If , then

So equality holds when we have or

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