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.

_________________ (\ /)
(O.o)
(> <)
This is Bunny. Copy Bunny into your signature to help him on his way to world domination

Joined: Mon, 29 Dec 2008 17:49:32 UTC Posts: 7276 Location: NCTS/TPE, Taiwan

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.

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

_________________ math puns are the first sine of madness -JDR

You cannot post new topics in this forum You cannot reply to topics in this forum You cannot edit your posts in this forum You cannot delete your posts in this forum