A beautiful inequality | International Mathematical Olympiad 2012 Problem 2

[@pradyumnanr3680]

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!

[@kanonmai]

Where are you, if it's in indonesia then that means you don't pass at pelatnas?

[@ethanyap8680]

Yay! One of the few questions you've shown I managed to solve :) Really love your videos!

[@mathcanbeeasy]

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]

And then 1+b(n)b(n+1)

[@mathcanbeeasy]

@@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). kzbin.info/www/bejne/mpDGY2uOed5ge7s

[@aaravrakesh4076]

Happy New Year! Keep posting, interesting stuff.

[@letsthinkcritically]

Happy new year!!

[@VerSalieri]

Nicely done, I would not have thought of that.. Keep up the good work.

[@letsthinkcritically]

Thank you!!

[@alexanderzxc1440]

Truly beautiful problem

[@johnnath4137]

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 really want to see your solution senpai. www.techaccents.com/2020/01/send-pictures-on-youtube-comments.html

[@johnnath4137]

@@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.

[@lqv3223]

@@johnnath4137 Amazing solution! Thanks for putting in the effort.

[@ImaginaryMdA]

Oh, I thought you were making a fermat's last theorem joke. :P

[@johnnath4137]

@@ImaginaryMdA Perhaps it was Fermat himself who made the joke and no one saw it for what it was for 400 years.

[@harshvadher9234]

I solved it by other method but still this holds totally strong than other methods I did with it.

[@cr1216]

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.

[@ShefsofProblemSolving]

Yeah the big idea was thinking of Am-Gm in the first place. It's a one creative idea problem.

[@Krish_202]

How about using weirstrass inquealitiy considering a2a3......an=1 but its not given that a2a3........an

[@mathematicalolympiad9157]

Can you make a video about IMO 2019 SL A2

[@letsthinkcritically]

I will look into that!!

[@amagilly]

Very nice techniques in this problem.

[@Maryam-Mehr]

Great problem, thank you!

[@吳沛洋-f8y]

elegant solution

[@BesnikDule-fl4yb]

I think was a very easy imo problem.

[@oo-sama3444]

Plz can u solve the imo p6 1988 I saw a lot of videos about it and i didnt understand it well

[@letsthinkcritically]

I will take a look at it!!

[@amirb715]

brilliant !

[@bernat8331]

Holder Inequality also kills the problem

[@VIKASVERMA-nk8uw]

Great

[@brumarul7481]

You are a fucking machine , jesus.

[@ИбадатЖұмабек]

nice

[@letsthinkcritically]

Thank you!!

[@VIKASVERMA-nk8uw]

Keep going

[@ГригорийЕ-в5ь]

Bruh what a2 a3 a4 and other can be except 1?

[@yuseifudo6075]

Anything really

[@soupayandasgupta8185]

Just beauty

[@gurjot5566]

Can you recommend me some books for mathematical Olympiad?

[@letsthinkcritically]

Sure!! I am going to share some Maths books that I enjoy in later posts

[@picontroversial308]

Grrrt

[@angelishify]

Zelo originalno.