Ah hmm that's what I feared, thanks for pointing that out.(Original post by Mladenov)
You are right, the points, at which the partial derivatives fail to exist, are good candidates for extremums. Yet, this does not help us at all.
I tried several methods to prove your inequalities, including some bruteforce techniques such as Lagrange Multipliers, but nothing works. Also, when , we have . Hence for all is clearly not true. I have not tried , but even if it is true, then you have to prove that is order your solution to work.
The problem is that you are doing the following:
Say you have correctly proven that for each , . Then, you are claiming that , which is generally not true.
Spoiler:ShowI recommend using some wellknown inequalities, in which case, the inequality is proven in less than a minute.
I suspected this was the solution you would have been after but found it difficult to arrange the terms in useful sets. Non the less, I shall persist!(Original post by Mladenov)
Precisely. There is a solution using exactly these two inequalities.

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Solution 73
I assume . Let be a finite multiset of positive integers. If all elements of are greater than , we know that the product of all the elements of is greater than or equal to their sum.
Suppose that we have solution with exactly components greater than . Thus the remaining components are equal to . Now, the sum of the elements, which are greater than , is less than or equal to the product; the sum of the 's is less than or equal to . It follows that .
As we are counting ordered sets, we have solutions in positive integers.
The number of nonnegative integer solutions of the equation is . Hence, the number of solutions in nonnegative integers is .
Problem 74 hint:
Last edited by Mladenov; 22042013 at 23:03. 
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 22042013 21:56
(Original post by Mladenov)
Solution 73
I assume . Let be a finite multiset of positive integers. If all elements of are greater than , we know that the product of all the elements of is greater than or equal than their sum.
Suppose that we have solution with exactly components greater than . Thus the remaining components are equal to . Now, the sum of the elements, which are greater than , is less than or equal to the product; the sum of the 's is less than or equal to . It follows that .
As we are counting ordered sets, we have solutions in positive integers.
The number of nonnegative integer solutions of the equation is . Hence, the number of solutions in nonnegative integers is .
Btw, I'm making some progress with your problem now usingSpoiler:which I wasn't familiar with until now (hopefully just a few algebraic tricks away from solution!)Showcs in Engel Form 
Lord of the Flies
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Solution 61
Assume continuous (perhaps this was not necessary).
Let , noting that . If for some , then for all over . However, then for , and hence over . Similarly if for some then over . In a similar vein, suppose there exist distinct such that then which implies and since we have over , which, as previously demonstrated, implies over . Similarly for the case . Hence if is not injective it is everywhere.
Naturally, we now assume is injective. Observe that but also that inductively, we have . Combining these facts, for positive integers . Hence we have . Now suppose that over then by the preceding statement, over and hence the inequality is satisfied over arbitrarily large intervals, which contradicts . Similarly if over any subinterval of . Hence we must have over . Now assuming over , our preceding proposition would imply that for and which is absurd. Similarly if over . Hence over as well.
Therefore or 
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(Original post by Lord of the Flies)
...
You are not using continuity, so it is not necessary.
Another approach:

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 23042013 19:02
(Still can't do problem 74 (fml!) so thought "if you can't beat them, join them")
Problem 75
Show that for positive real numbers a,b,c such that . 
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 23042013 21:20
Solution 75
Let , , .
We have .
Since , from
(AMGM), and
(obvious), we obtain
Spoiler:ShowThis is indeed nice application of Engel's form.Spoiler:ShowUse similar technique to prove 74. Without the substitution of course.Last edited by Mladenov; 23042013 at 22:41. 
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(Original post by joostan)
This may be a silly question but does anybody like geometry?? Even Mladenov who appears to be a maths genius dislikes it
Geometry is also taught extremely badly in most countries. 
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 24042013 00:30
But could you please explain how you have used AMGM in the first of the two inequalities?
And why is the second inequality obvious? It rests on and combination with the first result.
(sorry If this is actually obvious to you, but I don't find it to be so!)
Spoiler:ShowI stopped trying it yesterday but I might have another go tomorrow. What I've got so far is:Spoiler:If I'm miles off, a hint would be helpful, but if I'm along the right lines then just give me a nodShowBy using normalisation I deduced that abc=1 can be used WLOG. I then applied engel's form to get and I've fiddled around with it but haven't been able to find a way to nicely either algebraically combine it with the other art of the LHS, apply CZ, rearrangement or AMGM with the other part of the LHS, or find a way to apply one of these such a equalities to the expression itself such that it is greater than or equal to the difference of the RHS  *the other term of the LHS*. I've managed to get a few fairly simple expressions that the LHS is greater than but haven't quite hit the nail on the head.Last edited by Jkn; 24042013 at 00:31. 
ukdragon37
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(Original post by Llewellyn)
Geometry is wonderful, if you can do it
Geometry is also taught extremely badly in most countries. 
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Such a shame this thread is becoming a bit stagnant! All we need is 24 more problems before we can whip out the jayz jokes!
A problem people may recognise:
Problem 76*/**
For positive real numbers a,b,c, prove that
Last edited by Jkn; 24042013 at 19:49. 
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And another...
Problem 77*
Long John Silverman has captured a treasure map from AdamMcBones.
Adam has buried the treasure at the point (x,y) with integer coordinates (not necessarily positive).
He has indicated on themap the values of and , and these numbers are distinct.
Prove that Long John has to dig only in one place to find the treasure.Last edited by Jkn; 24042013 at 04:11. 
Lord of the Flies
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 24042013 04:43
(Original post by Mladenov)
Splendid solution.
You are not using continuity, so it is not necessary.
Another approach:
I should really be working, mais bon.
Solution 52
Note that since we have which then implies and inductively for any . Hence there must be some such that . Then clearly for all we must have where for any . But since is bounded is bounded as well, and hence we require for some . Thus , which, if plugged into the definition of immediately gives . Now we work backwards to show that all preceding terms are also . Obviously , and given that we clearly cannot have and hence it follows that etc. down to .
Therefore the only possible sequence is .
(Original post by Jkn)
But could you please explain how you have used AMGM in the first of the two inequalities?
And why is the second inequality obvious?
By the way, in Problem 75, notice that by plugging in a couple coefficients into the inequality and taking the exact same approach as in Mladenov's solution we can easily generalise the result:
Then using the same tricks to show that and we get:
For those how dislike inequalities with a passion:
Problem 78 (a classic!)
Evaluate
Problem 79
This is lolzmaths, but whatever. Let's see who can do it the fastest!
If findLast edited by Lord of the Flies; 24042013 at 06:28. 
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 24042013 10:28
(Original post by Jkn)
And another...
Problem 77*
Long John Silverman has captured a treasure map from AdamMcBones.
Adam has buried the treasure at the point (x,y) with integer coordinates (not necessarily positive).
He has indicated on themap the values of and , and these numbers are distinct.
Prove that Long John has to dig only in one place to find the treasure. 
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(Original post by Lord of the Flies)
Problem 79
This is lolzmaths, but whatever. Let's see who can do it the fastest!
If find 
Lord of the Flies
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 24042013 13:43
(Original post by james22)
Does f^(n) mean the nth derivative or f applied n times. I think its the former but if its the latter than answer is 0. 
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 24042013 17:41
(Original post by Jkn)
..
Problem 74 hint
Spoiler:Show
(Original post by Lord of the Flies)
Solution 52...
Solution 76
We can suppose , , and . AMGM and CS yield .
There are two possibilities: either my solution is totally wrong, or this is too weak inequality.
Solution 77
Suppose for the sake of contradiction that there are four distinct integers such that
;
.
, . Thus, . The case implies , . So, , contradiction. Let . We then have and therefore , . Consequently, contradiction. Hence our person (I do not remember his name ) has to dig in only one place.
Solution 78
Define . It is quite wellknown that . Hence .
There are several ways to solve this functional equation. We can bound above and below; or we can introduce the function and our expression is ; finally, we can consider the general equation , where , , and , , and solve it iteratively. 
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 24042013 18:23
(Original post by Lord of the Flies)
Problem 75, notice that by plugging in a couple coefficients into the inequality and taking the exact same approach as in Mladenov's solution we can easily generalise the result:
Then using the same tricks to show that and we get:
Holy crap, what an insight! The result looks so much more beautiful that way
Love this question! Kind of makes me wish I was born 100 years ago so I could've send a solution off to the magazine! (Ramanujan had to publish the solution himself I think. Deriving it from the more generalised version)Last edited by Jkn; 24042013 at 18:25. 
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Spoiler:than anything else (though you may find a different solution to me!) Dayyyyyum son you post fast!Showalgebraic tricks 
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Problem 76 is wrong in this form (I thought ).
Edit: We have to bound ; I can try to find the best possible .
Solution 79
As , from Taylor's series we have , .
Thus, .
Hence
Problem 80
Let be an increasing sequence of positive integers. Suppose also that . Then, the sequence contains all positive integers.Last edited by Mladenov; 24042013 at 19:59.
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