I never got to take part in an IMO, I couldn't attempt to qualify after my 9th grade. These videos help me get a feel for the quality of problems in IMO, so thanks a lot!
This is a very interesting problem for students who learn about induction. This is the case when the induction is about "the number of the numbers". So, in P(n) we have a product of n numbers =1 and in P(n+1) we have the product of n+1 numbers =1. This means we can't use the same numbers from P(n), because, if we use that numbers it means that a(n+1) and all others after that are 1. Example: P(3): 6*1/2*1/3=1 In P(4) if we use 6, 1/2, 1/3, than the fourth number must be 1. And so on. So, P(4) actually contain four other numbers than P(3). So, if in P(n) we have a1*a2*...*an=1 Then in P(n+1) we must have others b1*b2*...*bn*b(n+1)=1. So, how can we use P(n)? Simple by grouping 2 of that factors. c1=b1 c2=b2 ........... cn=bn*b(n+1) Of course, the hard part is the next part, when one of the factors (1+an) is now (1+bn*b(n+1))
@@aziz0x00 yes. Something like in this video. Is the best example of the induction where P(k) does not contain same operations or numbers like P(k+1). ua-cam.com/video/d_c28XDy-EY/v-deo.html
I’ve just concocted a nice proof of this result by induction. My solution has been MathTyped and I can only post it as a png or gif image, but You Tube comments slots do not support images, so I can’t upload it.
@@lqv3223 I tried to upload both my png and gif images as per your link, but was getting “incorrect format”, and gave up on this. I painfully copied out my image onto an MS Word document - it took ages. Here is my induction proof. Assume that for some n ≥ 3 and for some a₂, a₃,….,a(n) with a₂a₃….a(n) = 1, (1 + a₂)²(1 + a₃)³….(1 + a(n))ⁿ > nⁿ. I am now going to prove that the result is true for n + 1. This means that for any choice of (n + 1) a’s (not necessarily the same a’s as before, but in this case we’ll use the same a’s with a(n + 1) = 1) having the product property, the result for (n + 1) a’s holds. In particular, when I am proving the result for for n = 3, though I am going to use the symbols a₂, a₃ as before, they are not the same original a₂, a₃, but the new a₂, a₃ satisfy a₂a₃ = 1. Now (1 + a(n + 1) ⁿ⁺¹ ≥ (2√(a(n + 1))ⁿ2√(a(n + 1) = 2ⁿ⁺¹((a(n + 1)) ⁿ⁺¹ = 2ⁿ⁺¹ ( recall I have chosen a(n + 1) = 1). We have , (1 + a₂)²(1 + a₃)³….(1 + a(n))ⁿ(1 + a(n + 1)ⁿ⁺¹) > nⁿ2ⁿ⁺¹. Now (n + 1)ⁿ⁺¹ = nⁿ⁺¹((1 + 1/n) ⁿ⁺¹ < n ⁿ⁺¹((1 + 1/1)ⁿ⁺¹ (∵n ≥ 3) = nⁿ2ⁿ⁺¹ → nⁿ2ⁿ⁺¹ > (n + 1)ⁿ⁺¹ . So (1 + a₂)²(1 + a₃)³….(1 + a(n))ⁿ(1 + a(n + 1))ⁿ⁺¹ > nⁿ2 ⁿ⁺¹ > (n + 1)ⁿ⁺¹. The result is true for n = 3 ∵ (1 + a₂)²(1 + a₃)³ = (1 + a₂)²(1/2 + ½ + a₂)³ ≥ 2² a₂.3³(∛( a₃/2²))³ = 2² a₂.3³( a₃/2²) = 3³ (with a₂a₃ = 1). So by induction the result is true ∀n ∈ ℕ and ∀ a₂, a₃,….,a(n) with a₂a₃….a(n) = 1.
This seems unbelievably easy for an IMO problem. As soon as I see the word AM-GM I immediately came up with the solution even without writing anything down. Maybe connecting the problem with using AM-GM is the hard part.
I never got to take part in an IMO, I couldn't attempt to qualify after my 9th grade. These videos help me get a feel for the quality of problems in IMO, so thanks a lot!
Where are you, if it's in indonesia then that means you don't pass at pelatnas?
Yay! One of the few questions you've shown I managed to solve :)
Really love your videos!
Nicely done, I would not have thought of that.. Keep up the good work.
Thank you!!
Happy New Year! Keep posting, interesting stuff.
Happy new year!!
This is a very interesting problem for students who learn about induction.
This is the case when the induction is about "the number of the numbers".
So, in P(n) we have a product of n numbers =1 and in P(n+1) we have the product of n+1 numbers =1.
This means we can't use the same numbers from P(n), because, if we use that numbers it means that a(n+1) and all others after that are 1.
Example:
P(3): 6*1/2*1/3=1
In P(4) if we use 6, 1/2, 1/3, than the fourth number must be 1. And so on. So, P(4) actually contain four other numbers than P(3).
So, if in P(n) we have a1*a2*...*an=1
Then in P(n+1) we must have others
b1*b2*...*bn*b(n+1)=1.
So, how can we use P(n)?
Simple by grouping 2 of that factors.
c1=b1
c2=b2
...........
cn=bn*b(n+1)
Of course, the hard part is the next part, when one of the factors (1+an) is now (1+bn*b(n+1))
And then 1+b(n)b(n+1)
@@aziz0x00 yes. Something like in this video. Is the best example of the induction where P(k) does not contain same operations or numbers like P(k+1).
ua-cam.com/video/d_c28XDy-EY/v-deo.html
Truly beautiful problem
How about using weirstrass inquealitiy considering a2a3......an=1 but its not given that a2a3........an
I’ve just concocted a nice proof of this result by induction. My solution has been MathTyped and I can only post it as a png or gif image, but You Tube comments slots do not support images, so I can’t upload it.
I really want to see your solution senpai.
www.techaccents.com/2020/01/send-pictures-on-youtube-comments.html
@@lqv3223 I tried to upload both my png and gif images as per your link, but was getting “incorrect format”, and gave up on this. I painfully copied out my image onto an MS Word document - it took ages. Here is my induction proof.
Assume that for some n ≥ 3 and for some a₂, a₃,….,a(n) with a₂a₃….a(n) = 1, (1 + a₂)²(1 + a₃)³….(1 + a(n))ⁿ > nⁿ. I am now going to prove that the result is true for n + 1. This means that for any choice of (n + 1) a’s (not necessarily the same a’s as before, but in this case we’ll use the same a’s with a(n + 1) = 1) having the product property, the result for (n + 1) a’s holds. In particular, when I am proving the result for for n = 3, though I am going to use the symbols a₂, a₃ as before, they are not the same original a₂, a₃, but the new a₂, a₃ satisfy a₂a₃ = 1. Now (1 + a(n + 1) ⁿ⁺¹ ≥ (2√(a(n + 1))ⁿ2√(a(n + 1) =
2ⁿ⁺¹((a(n + 1)) ⁿ⁺¹ = 2ⁿ⁺¹ ( recall I have chosen a(n + 1) = 1). We have , (1 + a₂)²(1 + a₃)³….(1 + a(n))ⁿ(1 + a(n + 1)ⁿ⁺¹) > nⁿ2ⁿ⁺¹. Now (n + 1)ⁿ⁺¹ = nⁿ⁺¹((1 + 1/n) ⁿ⁺¹ < n ⁿ⁺¹((1 + 1/1)ⁿ⁺¹ (∵n ≥ 3) = nⁿ2ⁿ⁺¹ → nⁿ2ⁿ⁺¹ > (n + 1)ⁿ⁺¹ . So (1 + a₂)²(1 + a₃)³….(1 + a(n))ⁿ(1 + a(n + 1))ⁿ⁺¹ > nⁿ2 ⁿ⁺¹ >
(n + 1)ⁿ⁺¹. The result is true for n = 3 ∵ (1 + a₂)²(1 + a₃)³ = (1 + a₂)²(1/2 + ½ + a₂)³ ≥ 2² a₂.3³(∛( a₃/2²))³ = 2² a₂.3³( a₃/2²) = 3³ (with a₂a₃ = 1). So by induction the result is true ∀n ∈ ℕ and ∀ a₂, a₃,….,a(n) with a₂a₃….a(n) = 1.
@@johnnath4137 Amazing solution! Thanks for putting in the effort.
Oh, I thought you were making a fermat's last theorem joke. :P
@@ImaginaryMdA Perhaps it was Fermat himself who made the joke and no one saw it for what it was for 400 years.
This seems unbelievably easy for an IMO problem. As soon as I see the word AM-GM I immediately came up with the solution even without writing anything down. Maybe connecting the problem with using AM-GM is the hard part.
Yeah the big idea was thinking of Am-Gm in the first place. It's a one creative idea problem.
I solved it by other method but still this holds totally strong than other methods I did with it.
Can you make a video about IMO 2019 SL A2
I will look into that!!
Great problem, thank you!
Very nice techniques in this problem.
elegant solution
Plz can u solve the imo p6 1988
I saw a lot of videos about it and i didnt understand it well
I will take a look at it!!
Bruh what a2 a3 a4 and other can be except 1?
Anything really
brilliant !
I think was a very easy imo problem.
Holder Inequality also kills the problem
Great
Keep going
Can you recommend me some books for mathematical Olympiad?
Sure!! I am going to share some Maths books that I enjoy in later posts
Just beauty
nice
Thank you!!
You are a fucking machine , jesus.
Grrrt
Zelo originalno.