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# The Proof is Trivial! watch

As someone has said, it is .

How do you derive your first inequality? It is not true. And, just by the way, this inequality is quite sharp.
Oh gosh please spare me the embarrassment
2. (Original post by metaltron)
Spoiler:
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Solution 393

Squaring both sides:

Using the AM-GM we have:

Hence since a,b,c => 1:

Hence:

and we are done.

So either I have redeemed myself or dug an even bigger hole. I have my fingers crossed
Nope, it is still incorrect. The inequality does not hold.
Nope, it is still incorrect. The inequality does not hold.
Then I apologise for wasting your time with ridiculous solutions, hopefully I have woken up with a clearer mind today

Problem 394***

Find .
If you make the substitution, , then

This is simply equal to

However, this is not defined. Has something gone awry in the substitution?

Excellent problems, by the way!
5. This is the better alternative for 384, I'm pretty sure this was not mentioned in the official IMO solutions.

WLOG assume that

let

We define the infinite integer sequence :

By induction we can easily show

and

We must now to prove that one of the term must be

To prove this, let be the smallest positive term in the sequence such that , by definition that this term is smallest, we have

suppose that , then in the following equation (*), LHS is greater than 0 while RHS is less than or equal to 0, which is a contradiction

therefore , substitute into equation (*) we have
6. (Original post by henpen)
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If you make the substitution, , then

This is simply equal to

However, this is not defined. Has something gone awry in the substitution?

Excellent problems, by the way!
The substitution is fine and nice.

How did you obtain the mixed beta derivative?
7. (Original post by Felix Felicis)
Solution 388

Solution 389
It is very well known that (unless that was the intention of the question).

Alternative:

Solution 389

Consider . Choose some beautiful contour ; for example, one the most popular - square with vertices at will do. We can easily (indeed, all we need is to bound below) see that as .

Our function has simple poles at , and -pole at .
For , we have , which is obvious.

We turn to the case . We note and . Hence, .

Whence, .
Actually I had thought up the problem with the following solution in mind.

Alternative solution II to problem 389:

Consider the Fourier series

Where setting leads to the required solution.
The substitution is fine and nice.

How did you obtain the mixed beta derivative?

Let .
9. If you are having a boring Wednesday night, then play with to find solutions for .

For example, .
10. (Original post by henpen)
Spoiler:
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Let .
Well, we get the same thing without the substitution, so it does not give us any further information using it that way. I gave it some thought, and it seems that we can employ neutrices to evaluate this derivative (we get the correct result). If you are interested, I can provide some references.
Well, we get the same thing without the substitution, so it does not give us any further information using it that way. I gave it some thought, and it seems that we can employ neutrices to evaluate this derivative (we get the correct result). If you are interested, I can provide some references.
Very much so, could you quickly summarise what neutrices are? It seems counterintuitive that a function can have a defined derivative at a point where it is not defined.
12. Problem 395**

If

Show that
13. Solution 395 (Not exactly 3*s o.O)

(it doesn't really matter which variable you use...)

Switching to polar coordinates

...
I must leave now, if I've made any mistakes I'll correct on my return.
14. Hm, my method is similarish

I think most people have seen that question before, you guys should continue to post alternative methods
15. Nice. Glad i was able to do something...
(Original post by Flauta)
Hm, my method is similarish<br />
<br />
<br />
<br />
I think most people have seen that question before, you guys should continue to post alternative methods <img src="images/smilies/tongue.png" border="0" alt="" title="" smilieid="5" class="inlineimg" />
<br />
<br />
16. (Original post by keromedic)
Nice. Glad i was able to do something...
Me too, most of the questions on here are too hard for me to even attempt

(I tried to rep your method but it won't let me so I'll just try again later)
17. (Original post by Flauta)
Me too, most of the questions on here are too hard for me to even attempt

(I tried to rep your method but it won't let me so I'll just try again later)
I didn't realise that the question was fairly routine. I only post questions which I myself can do. Oh well

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18. (Original post by Arieisit)
I didn't realise that the question was fairly routine. I only post questions which I myself can do. Oh well

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It's still a really good question, there're some quite complex in-depth ways of doing it which I don't understand at all, would love to see some people post them. What was your method?
19. (Original post by Flauta)
Hm, my method is similarish

I think most people have seen that question before, you guys should continue to post alternative methods
Is there a method of solving this integral using just elementary methods (i.e. no double integrals or change of coordinate systems).
20. Another solution to 295:

Now, as and , we have:

Therefore

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