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# The Proof is Trivial! watch

1. (Original post by joostan)
Problem 72: *

Bit of an effort: - But hey ho.
Prove by considering: that

Find
Solution 72

After finding that

we get that

so

(the area must be above the x axis, and certainly positive).

For the next part, we have

Use

to get

Using partial fractions:

Substituting back gives

Simplifying further:

2. LoL OMG how did I end up here? Still at least I know it is possible to do equations on computers.
Can computers give you answers to those equations????
*probably a stupid question*
3. Solution 26

Notice that
As is a group, it follows that ,
Moreover, if , then, obviously, . It is easy to see that and are in , and therefore . Furthermore, implies . Trivially, . We can conclude that is a subgroup of .
Suppose that there is at least one element , which is not in .
The numbers and are relatively prime. Consequently, we can choose such that and .
Then, and . Hence, , which is clearly a contradiction.
You mean , don't you?
good spot
5. (Original post by Indeterminate)
Solution 72

After finding that

we get that

so

(the area must be above the x axis, and certainly positive).

For the next part, we have

Use

to get

Using partial fractions:

Substituting back gives

This simplfies a bit more I believe
6. (You'll probably all find this easy, but perhaps its a good one to make you think about initial approach. Plus i made it myself!)

Problem 73*

i) Find the number of ordered sets of positive integers such that and where m is a positive integer such that . Justify your answer.
ii) Extend your result to include all ordered sets of non-negative integers.
7. (Original post by joostan)
This simplfies a bit more I believe
If you insist, I'll edit it in
8. (Original post by metaltron)
Is that the simplest it can be? It looks like you can get to:

Spoiler:
Show
Surely

which isn't true in general.
9. (Original post by Indeterminate)
Surely

which isn't true in general.
No i deleted it, as I considered how inadequate my knowledge of the subject was having not looked at hyperboilc functions. It just seems to come straight from the line above that, but I'm probably missing something. How did you manage to reply to that, i thought I deleted it in the nick of time.
10. (Original post by metaltron)
No i deleted it, as I considered how inadequate my knowledge of the subject was having not looked at hyperboilc functions. It just seems to come straight from the line above that, but I'm probably missing something. How did you manage to reply to that, i thought I deleted it in the nick of time.
If you have a message quoted, I think it posts it regardless of whether the other person deletes it. When you delete a message, it only really removes it from view for the majority of the members of the forum, but it still has a unique post code and can still be quoted - so if someone quotes you before you delete it, it will have all the text you posted as well.
11. (Original post by Noble.)
If you have a message quoted, I think it posts it regardless of whether the other person deletes it. When you delete a message, it only really removes it from view for the majority of the members of the forum, but it still has a unique post code and can still be quoted - so if someone quotes you before you delete it, it will have all the text you posted as well.
Yeah, I know I've been humiliated many a time.
12. Problem 74**

Let be positive real numbers. Then we have the following inequality: .

Spoiler:
Show
It seems more difficult, than it actually is.
Problem 74**

Let be positive real numbers. Then we have the following inequality: .

Spoiler:
Show
It seems more difficult, than it actually is.
Spoiler:
Show

This method seems really trivial. I tried it on a few other inequalities and it always gave "min(LHS)=max(RHS)" (i.e: the solution) but I fear I have made some horrible assumption in applying it. Anyway here is my "solution"...

Noticing that either side on the equation is symmetrical in terms of the interchangeability of a, b and c, we can assume, W.L.O.G, that (*).

Simplifying we get: LHS=

As each term is positive and clearly the denominator of each term in less than or equal to by (*).
Simplifying we get... LHS for ALL a, b and c that satisfy (*).
The maximum value of this expression occurs when a=b=c (i.e: a+b+c=3c) so we get LHS (by the fact a=b=c had to be assumed in order to get the maximum).

Using the same technique, RHS. This clearly occurs when a=b=c also so RHS

Therefore LHSRHS.
The result follows.
14. (Original post by Jkn)
...
We have for all . Hence and equality occurs only when ?!

Edit: I mean, what makes you believe that is always true? You have for all , but does this imply for all ? Bear in mind that changes as you change .
We have for all . Hence and equality occurs only when ?!
I haven't got an upper bound for the LHS (I said it was greater that I think you misread!)
Edit: I mean, what makes you believe that is always true? You have for all , but does this imply for all ? Bear in mind that changes as you change .
Hmm I see your point. Well, short of using partial differentiation to justify this is a minimum, would it be sufficient to note that...
LHS by (*) and note that as a approaches c, each inequality approaches equality and hence demonstrate that the minimum value occurs at a=b=c? (if this works then the same could be easily done for the RHS). Cheers for the feedback btw, I wasn't sure it worked but I thought I'd give it go!
16. (Original post by Jkn)
I haven't got an upper bound for the LHS (I said it was greater that I think you misread!)
I gave you the upper bound just to make you think that you need to justify your inequalities.

(Original post by Jkn)
Hmm I see your point. Well, short of using partial differentiation to justify this is a minimum, would it be sufficient to note that...
LHS by (*) and note that as a approaches c, each inequality approaches equality and hence demonstrate that the minimum value occurs at a=b=c? (if this works then the same could be easily done for the RHS). Cheers for the feedback btw, I wasn't sure it worked but I thought I'd give it go!
Your argument proves only that , when . Also, should it not be ?

If we denote , then we have to prove that there are no such that ; in other words, we have to prove that reaches its minimum when .

Partial differentiation would work, but it is quite tough task.
I gave you the upper bound just to make you think that you need to justify your inequalities.

Your argument proves only that , when . Also, should it not be ?
Sorry, I'll change that now
If we denote , then we have to prove that there are no such that ; in other words, we have to prove that reaches its minimum when .

Partial differentiation would work, but it is quite tough task.
Well using partial derivatives on the 6-term expression with respect to a, b and c in turn we get that...
which implies f(a,b,c) approaches a stationary point as all three variables coalesce to zero. Choosing a few values above and below shows you that this is a minimum point and hence, the minima occurs when and only when a=b=c.
18. (Original post by Jkn)
Well using partial derivatives on the 6-term expression with respect to a, b and c in turn we get that...
which implies f(a,b,c) approaches a stationary point as all three variables coalesce to zero. Choosing a few values above and below shows you that this is a minimum point and hence, the minima occurs when and only when a=b=c.
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 brute-force 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:
Show
I recommend using some well-known inequalities, in which case, the inequality is proven in less than a minute.
Spoiler:
Show
I recommend using some well-known inequalities, in which case, the inequality is proven in less than a minute.
Which ones? I know cauchy-schwarz and AM-GM inequalities, is that enough?
20. (Original post by james22)
Which ones? I know cauchy-schwarz and AM-GM inequalities, is that enough?
Precisely. There is a solution using exactly these two inequalities.

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