I usually skip inequalities...

[@demenion3521]

for the second part of the problem, you can see that 1/r*(1-r^n)/(1-r) < 1/(r-r²) ≤ 4 --> r²-r+1/4 ≤ 0. the only possible solution to that is r=1/2, so that x_n=1/2^n is the only possible geometric sequence that solves this problem. EDIT: in case anyone else was wondering, for a harmonic sequence x_n=1/(n+1) you get N=7 as smallest value for N.

[@goodplacetostop2973]

14:25

[@mikeschieffer2644]

At 1:45 mentioning the Cauchy-Schwarz Inequality on May 4th. Nice! May the Schwarz be with you!

[@obnoxious2471]

This application of Cauchy Schwartz is also known as Titus Inequality

[@alexismiller2349]

In France we sometimes call it the inequality "des mauvais élèves" literally meaning, the bad students' inequality. As in when you try to simplify the sum of fractions on the left hand side, you somehow (by really messing it up) get the expression on the right hand side. :)

[@TechToppers]

@Ꮶαƚѕuƙi βαƙugօ Tons more... Muirhead Schur Jensen Bla bla

[@goodplacetostop2973]

@@alexismiller2349 Bad students' inequality... That's interesting, it must have some irony behind that name 😛

[@TechToppers]

@@goodplacetostop2973 I heard this first time. I called it Titu's Lemma as everyone else.

[@alexismiller2349]

@Ꮶαƚѕuƙi βαƙugօ ​ adding to Tech Toppers there is Minkowsky's inequality, Chebychev's inequality, the rearrangement inequality, the generalized weighted mean inequality as well as the Maclaurin and Newton's inequalities as well as Karatama's inequality. I think that covers almost all the inequalities that you could find in math competitions like the IMO. Except if you take into account geometry, graph theory, extremal combinatorics since those have there own set of inequalities :) (edit: totally forgot Hölder's inequality, which actually generalizes Cauchy Schwartz)

[@hydra147147]

This modify inequality is sometimes called C-S inequality in Engel form, Sedrakyan's inequality or Titu's lemma and comes up ridiculously often in Olympiad Math.

[@saikat93ify]

Yes, but even after knowing that I find it difficult to apply it to a new problem !

[@Hiltok]

Nailing down the solution to the second part ... To nail down why 1/2 is the solution to the second part: Take the limit as n -> inf of (1/r)(1-r^n)/(1-r) which is (1/r)1/(1-r)=1/(r-r^2). Taking the derivative of this w.r.t. r gives (2r-1)/(r-r^2)^2, which is zero exactly when r=1/2. This is the local minimum on (0,1) where 1/(r-r^2) is strictly positive.

[@blazedinfernape886]

I usually skip inequalities.... But makes inequalities in every single math contest video.....

[@yassir641]

Haha, True

[@jakubkocak887]

Just little insight why to choose geometric sequence in the second part. You want a case, the first inequality is equal. And Cauchy-Schwarz inequality (≥) turns into equality (=) for colinear sequences. Therefore x_(i-1)/sqrt(x_i) = A*sqrt(x_i) for all i. Thus x_(i-1) = A*x_i. And we have geometric sequence. Now you just have to find correct parameter r.

[@littlefermat]

The moral: Seeing squares in the numerators of an inequality --> Titu's lemma saves the day

[@TechToppers]

It is ridiculously intuitive. This is the best part.

[@failsmichael2542]

Imagine you are a fresh contestant in 1982.

[@TechToppers]

@@failsmichael2542 Well, times have changed😂. It might have been a difficult problem back then...

[@HideyukiWatanabe]

I think it's one of the easiest problem in IMO. 5:23 Setting S=x_1+...+x_n looks easier; RHS = S+2(1-x_n) + (1-x_n)^2/S >= 4(1-x_n) by AMGM. Taking N s.t. x_n < 400 for all n>=N completes the proof of convergent case.

[@hugohugo37]

A guy named "little fermat" has some nice contest prep videos on here that this audience would like. His channel is new and small but his videos are well explained. (And no, I'm not him. I just stumbled across his channel.)

[@nebulasy8]

My day doesn’t start unless I’ve seen your 8 o’clock video.

[@austinm271]

You added an extra 9 (3.999 vs. 3.99 in the problem definition) - the right answer is probably n = 400 rather than n = 4000.

[@demenion3521]

well, 3.999 > 3.99, so technically he still solved the problem, just with a tighter bound. also N=400 is not necessarily the smallest value of N such that the inequality holds, but from this point on you can guarantee with this proof that the inequality is satisfied

[@TechToppers]

@@demenion3521 The original IMO problem has 3.999. See this www.imo-official.org/year_info.aspx?year=1982

[@tajpa100]

you are very brilliant and your lectures are true masterpieces

[@goodplacetostop2973]

I really like the Guts Round from HMMT, so here’s more from that competition. This time it’s from 2004 HOMEWORK : You want to arrange the numbers 1, 2, 3, ..., 25 in a sequence with the following property: if n is divisible by m, then the nth number is divisible by the mth number. How many such sequences are there?

[@goodplacetostop2973]

@David Schmitz SOLUTION *24* Let the rearranged numbers be a_1, ..., a_25. The number of pairs (n, m) with n|m must equal the number of pairs with a_n|a_m, but since each pair of the former type is also of the latter type, the converse must be true as well. Thus, n|m if and only if a_n|a_m. Now for each n= 1, 2, ...,6, the number of values divisible by n uniquely determines n, so n = a_n. Similarly, 7, 8 must either be kept fixed by the rearrangement or interchanged, because they are the only values that divide exactly 2 other numbers in the sequence; since 7 is prime and 8 is not, we conclude they are kept fixed. Then we can easily check by induction that n = a_n for all larger composite numbers n ≤ 25 (by using m = a_m for all proper factors m of n) and n = 11 (because it is the only prime that divides exactly 1 other number). So we have only the primes n = 13, 17, 19, 23 left to rearrange, and it is easily seen that these can be permuted arbitrarily, leaving 4! possible orderings altogether.

[@mathematicalminds8446]

Waao this serms really a hardcore proof and technic used in this video,, thanks for this video

[@supnava8320]

very elegant solution :) also please share your diet and workout routine, you look great!

[@shresthshukla6239]

Watched a random video in recommendation and continuously watching your videos. This one is 5th video in a row.🤘🤘

[@gmchess7367]

Very intresting problem👍👍

[@andreypillay4323]

I would really recommend the Cauchy Schwartz masterclass book for getting into inequalities

[@MrCoffeypaul]

Inequalities and Voodo go hand in hand!

[@mattattack57]

Loved this problem!

[@MathElite]

Nice video!! nice identity

[@timurpryadilin8830]

very nice solution!

[@AnkhArcRod]

I don't think any other choice except r= 1/2 would have worked for the second part!

[@tieliu6705]

From 5:44 to 8:19, you were really just saying (1+S)^2 \geq 4S in the numerator, but that should be just a one-liner, right?

[@robertgerbicz]

Exactly, proved that (1+s)^2/(s+x_n)>=4/(1+x_n/s) what is (1+s)^2>=4*s, but it is a triviality: (1-s)^2>=0.

[@h4ck314]

Minor typo: it's Schwarz, not Schwartz.

[@TechToppers]

😂 Someone noticed it...

[@allykid4720]

Say that to "borschtch".

[@goodplacetostart9099]

Keep starting at 0:01

[@leif1075]

Surely you can solve this without Caichy Schwartz since a lotnof3people arent going to think of that..say by replacing the numerator of each term with the denomiantor plus some quantity to get 1s in the sequence or plugging in different values and trying them?

[@242math]

excellent work, this was a 'nice' inequality bro

[@Mathcambo]

Nice solution

[@arandomcube3540]

Don't know why, but that's one of the most impressive video I've ever seen.

[@idjles]

chalk skills

[@audience2]

Watch more of his videos. That's actually pretty standard fare.

[@arandomcube3540]

@@audience2 Lol I've watched his videos for a year, right now

[@bprpfast]

First!!!

[@MusicIsCool-ll4ry]

Sorry bprp fast, you're second :(

[@goodplacetostop2973]

@@MusicIsCool-ll4ry 😎

[@piotrfelix]

That was really fast. Isn't a night in the US right now?

[@MusicIsCool-ll4ry]

@@piotrfelix It's the start of the morning for me

[@piotrfelix]

@@MusicIsCool-ll4ry So the East Coast :P

[@mrmanning6098]

I thought Michael was cool until I saw him using curly brackets { } for sequences... a type of naturally ordered set. I know this is tradition at some schools, but shouldn't we be using parenthesis or square brackets if we are talking about a set that has order?

[@ameerunbegum7525]

I don't know for sure but it seems that the volume of the video is a bit low.

[@TechToppers]

This video could be of 5-6 minutes instead of 14 mins. That Cauchy result is well known as Titu's Lemma. Then for applying AM-GM, you don't need that manipulation. Enough experience with this stuff, we know that 1+s^2>=2s and as this is inequality problem, why not AM-GM? Rest is doable ig. Btw, its his wish to present in this fashion. Not against it, just writing it all here.

[@awesomechannel7713]

In problem 1 how S is defined if n=1? If it's zero, you cannot just divide by it. He should have worked around that just saying that we will always pick N>1 since if inequality holds for some N it will hold for all bigger values of N and with that regard S is well defined.

[@ConManAU]

That’s a fair concern and I think it should be put in as a condition, but since the target n was much larger than 2 it doesn’t affect the solution.

[@dertyp6074]

But on left side of the Board it says that x(of 0)=1 so you would have 1 but it also says 0

[@ConManAU]

@@dertyp6074 Since s is the sum of the terms from x_1 to x_(n-1), if n = 1 then s is the sum of zero terms and so is zero.

[@dertyp6074]

@@ConManAU Tanks I mistook it as a sum of x_n-1 which it is obviously not. You are right of course

[@awesomechannel7713]

@@dertyp6074 I'm not sure. If not stated explicitly or not implied by initial conditions I'd assume N=1 is possible. The initial sum is generally written with more than 2 arguments to make pattern easier to recognize yet everyone understands there may be edge cases where sum consists of one summand. Anyways, point of my comment was tofind even a tiny inaccuracy and ruthlessly criticize for it.

[@jbtechcon7434]

Social unrest is caused by inequalities.

[@kouverbingham5997]

please make a video about how we dont know whether pi+e is rational or not, and how pathetic that is

[@pwmiles56]

Alternatively: claim that the result of Part 2 i.e. x(n)=1/2^n, n=0 to inf., gives the minimum infinite sum of the specified form, i.e. value 4. If this is proved then Part 1 follows [by the Bolzano-Weierstrass theorem, essentially guaranteeing the truncated series approaches arbitrarily close to its limit]. So: put {x(n)} = {1, a/2, ab/4, abc/8, ...} with a, b, c... all real and greater than 0. The minimal sum, S0 say, has a=b=c= ... =1. The general sum can be written as nested brackets, S = 2/a + a(1/b + b(1/2/c + c(1/d/4 + ... ))). Initially put b=c=d= .. = 1. It is easy to show that >>this>this

[@jeanf6295]

I like your approach, here is my take on it : For a sequence of length N+1, let's write x(n) = a(N-n)x(n-1) for 1