A MEAN Problem from India [ 2016 RMO Mathematical Olympiad ]

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A MEAN Problem from India [ 2016 RMO Mathematical Olympiad ]

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Күн бұрын

Wanna prepare yourself for a mathematics competition or just feel like trying out new & interesting problems? Why not try out Brilliant then? =D brilliant.org/...
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Today we'll be using the AM GM inequality to find out an upper bound for the product of 3 unknows given a fractal equation :) Enjoy! =D Video sponsored by Brilliant btw :3
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Пікірлер: 338

@kathanshah8305 3 жыл бұрын

Thumbnail and title suggests that fresh toadwalker has inspired you

@PapaFlammy69 3 жыл бұрын

nah

@3ckitani 3 жыл бұрын

I think it's from Michael Penn

@mikaeelshah7205 3 жыл бұрын

FReSh ToAdWalKeR: The answer is TOO good. a GOLDEN puzzle. TrAp (Trapezium)

@kathanshah8305 3 жыл бұрын

@@3ckitani yeah mixed it up

@kathanshah8305 3 жыл бұрын

@@PapaFlammy69 yeah it’s inspired by micheal penn not mindyourdecision

@AndrewDotsonvideos 3 жыл бұрын

This changes everything.

@kshitijgarg1470 3 жыл бұрын

Sir that move of dividing the both sides by 4 was dope 🔥. Love from India.

@TechToppers 3 жыл бұрын

You don't need that... AM GM destroys it on the spot.

@kshitijgarg1470 3 жыл бұрын

@@TechToppers Ya that's the same thing

@rajnikantsharma4054 3 жыл бұрын

@@TechToppers the main matter was that this man has to show power of Indian competitions

@kathanshah8305 3 жыл бұрын

Don’t lie you forgot password of flammable maths 2 so you posted this video here

@JonathonRiddell 3 жыл бұрын

Embarrassed to say I did not know the AM GM inequality

@bsmith6276 3 жыл бұрын

Embarassed to say I do know AM/GM inequality but still forget to ever use it.

@adityaruplaha 3 жыл бұрын

@@bsmith6276 same man

@devjitghosh550 2 жыл бұрын

so do i bro

@1987Videolover 2 жыл бұрын

AM GM i think never told to a reguler student...

@muahmuah4135 3 жыл бұрын

There are not much people who love mathematics and usually dedicate an entire channel and talk about it with such passion...... except for those teachers who are OP because it really hard and honestly nobody give a shit... so i really appreciate guys like you who has great passion for the subject

@AlphaCurveMath 3 жыл бұрын

As an Indian I can only warn you against the sheer horrors of the JEE Advanced content train. The audience is half a billion whiners who will hear nothing but praise for India and how the rest of the world would be stumped by JEE. It's only a downward spiral from there which, sadly enough, one of the potentially best physics channels has already headed into. Edit: But the olympiads are fine. It's about the only place where actual math problems come up. It's the jee's that are subpar.

@PapaFlammy69 3 жыл бұрын

ye, that is why I mostly tefuse to cover indian things :/

@aradhya_purohit 3 жыл бұрын

@@PapaFlammy69 papa, how do you 'tefuse'?

@AlphaCurveMath 3 жыл бұрын

@@PapaFlammy69 yeah if only to keep memeing as hard as you do right now, you've got to keep away from those who can't take a joke.

@anmoldeepsingh9281 3 жыл бұрын

Holy shitplus jee is overrated

@aradhya_purohit 3 жыл бұрын

@@AlphaCurveMath seconded.

@prashant2650 3 жыл бұрын

Ur Indian impression is like what we do for Bri-ish peeps

@theblinkingbrownie4654 3 жыл бұрын

Apparently you're 'better' than other countries' people if you can memorize and solve equations, even if you don't get any of the intuition, I live close to India and hate it when my friends brag about 'knowing' math. :/

@PapaFlammy69 3 жыл бұрын

...

@adityaruplaha 3 жыл бұрын

As an Indian, here competition is tough enough that it's the only way you can get an advantage.

@einzelganger7744 3 жыл бұрын

This is the first level examination. Even IMO problems involve AM GM and other inequalities. I've seen level 1 question papers of most countries and tbh they are all kinda same. Try INMO problems.

@theblinkingbrownie4654 3 жыл бұрын

@@einzelganger7744 you talking to me?

@seriouslysupersonic 2 жыл бұрын

What's up with all that violence? Why do the terms on both sides all have to "die"? Can't they just hapilly retire to the infinitesimal garden where all the epsilons and deltas hang out?

@PapaFlammy69 2 жыл бұрын

nah, they all ded

@alokmishra1414 3 жыл бұрын

7:33 yeah the education system really sucks here..it never forced our brain to think out of the box..instead always to remember tricks and formula on which we rely..i remember solving these questions not even knowing what i am actually doing..just do it man..just do it don't ask why and how..😓

@PapaFlammy69 3 жыл бұрын

exactly

@supercuntt5876 3 жыл бұрын

Is this sad that I was taught this kinda trick when learning about inequalities? Lol

@sohamvirkar802 3 жыл бұрын

Honest replies are deleted here!😂

@sohamvirkar802 3 жыл бұрын

Time to find new teachers bro😂 dont blame the whole education system just because YOU are taught that way

@supercuntt5876 3 жыл бұрын

@@sohamvirkar802 lol ok

@rudeus6621 3 жыл бұрын

me who qualified rmo in 2018 : 👁👄👁

@Prodbybah 3 жыл бұрын

bhai please say how and which books and which important chapters?

@aniketeuler6443 3 жыл бұрын

Now Jens is making Indians feel good 😂

@akshat9282 3 жыл бұрын

YOU HAVE NO IDEA HOW MANY VIEWS THIS IS GONNA GET JUST FOR HAVING INDIA IN ITS NAME. we 1.3billion indian normies love nothing more than seeing foreigners talk about our country

@onkarsingh8358 3 жыл бұрын

Right😂

@daphenomenalz4100 3 жыл бұрын

🤣

@sachinnandakumar1008 2 жыл бұрын

Yeah that is why crappy indian reaction channels like jaby koay has millions of subscribers

@merlinpinkfeather 2 жыл бұрын

Uses the word “normie”. Cringe

@chaitalichatterjee4742 Жыл бұрын

@@sachinnandakumar1008 cringey

@IshanBanerjee 3 жыл бұрын

There is a reason why I love RMO so much

@gamingsquad7149 Жыл бұрын

Ayo ishan saar here! 💀

@keshavb3128 3 жыл бұрын

In Papa Flammy's perspective, he trapped Andrew in his basement. In Andrew's perspective, he trapped Papa Flammy in his attic. You've been trapped in Andrew's attic, Papa Flammy.

@She_898 3 жыл бұрын

It's really sad here in India every high school kid is just trying to crack exams whether it is JEE or RMO or anything like that but no one is really enjoying what they are doing. I wish that students had some amusement about learning maths and physics rather than just clearing this exams

@mathsman5219 2 жыл бұрын

If you really love Mathematics ❤️ You will never want to check yourself through any exams (as they can't .)

@kumaracademy7555 3 жыл бұрын

7:17 I always heard that Germans are really honest..Today I saw....xD

@MathElite 3 жыл бұрын

Love pro gamer moves!!!!

@nandansantosh5497 3 жыл бұрын

Jokes on you man this was taught to us as joke by our teacher because we had a lot of time left on our hours

@Jop_pop 3 жыл бұрын

Alternative solution with lagrange multipliers: Substitute x=(a+1)/a, y=(b+1)/b, and z=(c+1)/c. We can define g(x,y,z)=1/x+1/y+1/z=1 (given) and f(x,y,z)=x+y+z. Very straightforward method of Lagrange multipliers then gives x=y=z=3. This can be verified to be the min for f. Thus xyz+x+y+z >= xyz+9. Next, notice that since 1/x+1/y+1/z=1, multiplying by xyz easily shows xyz=xy+yz+xz. So xyz+x+y+z >= xy+yz+xz + 9, thus xyz-(xy+yz+xz)+x+y+z-1=8. And so (x-1)(y-1)(z-1)>=8. Hence 1/(abc)>=8 and abc

@gamingstars8956 3 жыл бұрын

In most of inequalities just equate all terms to get min value a/1+a=b/1+b=c/1+c to get a=b=c=1/2

@lauthomas7179 3 жыл бұрын

abc attains its maximum only when a=b=c. So 3a/(a+1) =1 3a - a =1 a. =1/2 Max of abc = 1/8 Or abc=

@veralgupta8182 3 жыл бұрын

Oh yes my mentor is here with the question I failed to answer

@ishansingh2391 3 жыл бұрын

Try geometry problems of mathematic olympiads as RMO is 1st stage of selection process in Olympiad. obviously some of these problems may be easy but in our Olympiads even after writing your solution one may not get a whole 17/17 he may get 13 or 14 marks as the solution requires elegance not rigor to be precise.

@zh84 3 жыл бұрын

Now solve this simple problem: a/(b + c) = b/(c + a) = c/(a + b), a, b, c all positive integers.

@RC32Smiths01 3 жыл бұрын

Always enjoy these kinds of videos man! Awesome work

@karthikboyareddygari568 3 жыл бұрын

This just showed up in my recommended, and I gave it a shot. I ended up doing what you did up until you divided by 4 because I was not aware of that relationship between the arithmetic and geometric means. Not having that knowledge, I treated it as an optimization problem now that the constraint wasn't so hairy and used Lagrange multipliers to conclude that a = b = c = 1/2 was a point of interest. Seeing as a, b, c \in (0,\infty), there weren't really any other points to check. Using the value of 1/2, we find that abc = 1/8, which is now either an absolute max or min. By using a = b = 1/3 and c = 1 as a random point that satisfies the constraint, we obtain abc = 1/9 < 1/8; therefore, abc \leq 1/8 for all positive real a, b, c subject to the constraint.

@cr1216 3 жыл бұрын

Here is a way with no complex calculation at all. Let a=tan^2(x), b=tan^2(y), c=tan^2(z) then we get sin^2(x)+sin^2(y)+sin^2(z)=1, notice that cos^2(z)+sin^2(z)=1 by definition so this means cos^2(z)=sin^2(x)+sin^2(y) >= 2sin(x)sin(y) (last step by AM-GM) . Similarly we get cos^2(y) >= 2sin(x)sin(z) and cos^2(x)>=2sin(y)sin(z). Multiply the three together we get cos^2(x)cos^2(y)cos^2(z)>=8sin^2(x)sin^2(y)sin^2(z) and therefore 1>=8tan^2(x)tan^2(y)tan^2(z) which implies 1>=8abc.

@anuragsamanta9016 Жыл бұрын

Take c/1+c to RHS .then apply AM, GM in between a/a+1 and b/b+1...After getting three such inequality in ab bc and ac multiply all three equation. You will get the desired result

@udbhav5079 3 жыл бұрын

RMO simps for AM > GM > HM...

@PapaFlammy69 3 жыл бұрын

ye lol

@viinisaari 3 жыл бұрын

6:25 It's aRithmetic for the noun and arithMetic for the adjective.

@dijkstra4678 2 жыл бұрын

I was just thinking that the problem seemed pretty symmetric and then it hit me... what if we assume a=b=c? that would mean that a/(1+a) = 1/3 which means a = 1/2 which would then mean that abc = 1/8 proving abc

@DhirajKumar-rx8hi 2 жыл бұрын

Just Apply two times AM GM inequality. First for the given 3 terms a/(a+1). Product of abc/(a+1)(b+1)(c+1)

@harshpandey6725 3 жыл бұрын

Love from India🇮🇳 Really well solved!! Loved the pro gamer moves❤

@paulm5441 3 жыл бұрын

After reaching to 2abc + ab + ac + bc = 1 you can just plug it into the obvious (a+b+c)^2 >=0 . No need for AM, GM etc.

@AryanSingh-ro4fi 2 жыл бұрын

So satisfying

@doodelay 2 жыл бұрын

dividing by 4 and introducing the AM/GM inequality haha keep showing us the way papa flam (PS, most of my mathematical applications comes from watching u work in real time)

@markgh2491 3 жыл бұрын

I think it’s pronounced arithmetic, not arithmetic.

@PapaFlammy69 3 жыл бұрын

yeye

@hsiochen7216 2 жыл бұрын

I believe given two functions, f(a, b, c) and g(a, b, c), if both f and g remain the same after reordering their argument a, b, and c. g(a, b, c) will have a local maximum or minimum at the point of a=b=c, under the restriction f(a, b, c) = constant. A lot of these types of problem comes from function with symmetric argument.

@sisyphus645 3 жыл бұрын

The meme before the math is like the prayer before the food

@jmof0464 3 жыл бұрын

people that do those problems in india always endup in tech support

@ramanunnikrishnan7354 3 жыл бұрын

finally RMO problems, never got to give it

@mayankchauhan2814 3 жыл бұрын

This question is used as a AM class illustration in my math class

@chaitanyavarma1747 3 жыл бұрын

6:29 Lmao. Also first guy to comment is an Indian XD

@californium-2526 3 жыл бұрын

That NaCl meme, I know it...! A zillion reposts and copies of this (low-quality) meme has been seen, mainly in r/chemistrymemes.

@farhannr28 3 жыл бұрын

Bruh i literally just did that question on a test lmao. Lucky that i found the correct answer in last second

@rohitkhatri3892 3 жыл бұрын

7:18 I felt that.

@tobiasgorgen7592 3 жыл бұрын

It takes indian youtube to teach the entire world mathmatics. But it takes a pasty white boi from DEUTSCHLAND to school indians. I'm gonna tell my son that this is the circle of life.

@piyushpathak5012 3 жыл бұрын

And to school you

@tobiasgorgen7592 3 жыл бұрын

@@piyushpathak5012 Could you repeat that in brail, I can't read.

@piyushpathak5012 3 жыл бұрын

@@tobiasgorgen7592 Then get the f outta here

@gamingstars8956 3 жыл бұрын

I m watching maths for fun not for schooling uncle

@piyushpathak5012 3 жыл бұрын

@@gamingstars8956 Same

@blaiseragon8142 3 жыл бұрын

Love your content so far !

@PapaFlammy69 3 жыл бұрын

thx :)

@butter5014 3 жыл бұрын

Dear Papa Flammy (capital p because now you are deified just like Santa Claus), please help me out with the following problem from one of the Bulgarian mathematical olympiads: A hotel has 26 rooms. Each room has a number from 1 to 26. Different rooms have different numbers. Two guests are placed in rooms x and y. If the product xy is equal to the sum of the numbers of all rooms except x and y, find |x-y|. You are allowed only a pen and some paper. Thanks in advance!

@paulm5441 3 жыл бұрын

You have x, y between 1...26 and xy = 1+2+3+...+26 - x- y xy = 351 - x- y xy +x+y = 351 x(y+1) + y = 351 x(y+1) + y+1 = 351 + 1 (x+1)(y+1) = 352. 352 = 11*2^5 and the only product of two numbers x=1 and y+1 equalling 352 and having x+1 and y+1 in 2...27 is 22*16. So x+1 = 22 and y+1 = 16 or x+1 = 16 and y+1 = 22, giving us x= 15 and y = 21or vice-versa. It doesn't matter since |x-y| is the same, namely 21-15 = 6

@butter5014 3 жыл бұрын

@@paulm5441 Thanks for that! I hope you enjoyed this problem.

@tvbbian 3 жыл бұрын

If you let a=(tanA)^2, b=(tanB)^2, c=(tanC)^2, 0

@faenzarfaenzar2636 3 жыл бұрын

Amazing video ! But I got one question : Where did you get that Blackboard !!!!!???????? I want one !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

@chrisglosser7318 3 жыл бұрын

So... why didn’t you just use Lagrange multipliers to show that the constrained abc have a maximum when they are equal(?)

@adityaekbote8498 3 жыл бұрын

@Chris Glosser what are Lagrange multipliers??

@chrisglosser7318 3 жыл бұрын

@@adityaekbote8498 They are a way implementing constraints in variational calculus

@adityaekbote8498 3 жыл бұрын

@@chrisglosser7318 ohh thank you

@yoyokojo651 3 жыл бұрын

Kinda tedious tbh, gotta argue compactness of constraint ect. Plus it’s kinda a lot of Algebra in this case

@BenjusJamentus 3 жыл бұрын

Is this closer to the kind of mathematics that is done in university or that we do at the "Gymnasium" (some kind of preparation school for university in Switzerland/Germany/Austria I think as well)? I'm really interested in studying maths but I'm not sure what to expect... Greetings from Switzerland!

@PapaFlammy69 3 жыл бұрын

something in between I would say^^

@BenjusJamentus 3 жыл бұрын

@@PapaFlammy69 OK, thank you!

@timburdack7366 3 жыл бұрын

Love this! ❤

@avtarsingh-tz1up 3 жыл бұрын

India❤❤

@vidhanp482 3 жыл бұрын

Micheal Penn cameo

@MrRyanroberson1 3 жыл бұрын

notice: if we substitute x,y,z = 1/a,1/b,1/c, the problem becomes the following: 1/(1+x) + 1/(1+y) + 1/(1+z) = 1 -> xyz >= 8. therefore let us multiply all by (1+x)(1+y)(1+z) to get 3 = 1+x+y+z+xy+yz+xz+xyz. so on, so forth

@MrRyanroberson1 3 жыл бұрын

why am i like this - i made such an elementary mistake. fixing time: we get (1+x)(1+y) + (1+x)(1+z) ... and it doesn't get all that much simpler, so this substitution doesn't help much besides the initially simpler representation

@PapaFlammy69 3 жыл бұрын

happens to the best, Ryan

@ShefsofProblemSolving 3 жыл бұрын

Man that division by 4 was so out of left field for me. Also you're a beast for actually multiplying this out. A more natural approach for me was let x = (a+1)/a, y = (b+1)/b, z = (c+1)/c. x, y, z > 1 The condition is 1/x + 1/y + 1/z = 1 and the problem asks us to prove that (x-1)(y-1)(z-1) => 8 which is true ifff x + y + z => 9 (xy + yz + zx = xyz by condition). Now this is just AM - HM because (x+y+z)/3 => 3/(1/x + 1/y + 1/z) = 3. Q. E. D. Baby

@PapaFlammy69 3 жыл бұрын

@ShefsofProblemSolving 3 жыл бұрын

@@PapaFlammy69 He replies ! (let me think of something interesting to say... uh "I don't need to find x because I found you" - Nah too direct, I'll go with the cool option...) Hello there chap

@Maths_3.1415 Жыл бұрын

@@ShefsofProblemSolving 2 years late but hi

@mohamedababou8777 3 жыл бұрын

It is better if this is the question at world Mathematics conferences and Olympiad: "Those who believe that Numbers have no end are required to prove infinity realistically and concretely."

@latinirba4953 3 жыл бұрын

I have followed your channel since it was small and recently I saw you got 7 million views. I did not expect a maths channel to blow up. I hope you never attempt a JEE question, schoolkids should stop asking every science youtube channel to solve JEE questions. 😑 You won't be able to prepare for that exam by watching one or two problems solved anyways.

@quickyummy8120 3 жыл бұрын

I found another amazing Olympiad video kzfaq.info/get/bejne/hsCeqNd1vZaWo30.html

@qq-rf6hv 3 жыл бұрын

this is a really good solution! but I think there is way to make it easier, by using the fact that the product of numbers whose sum is constant is maximal when these numbers are equal

@karthikboyareddygari568 3 жыл бұрын

I was not aware of this fact, though I'm sure it's easy enough to show as a matter of optimization. In this case, all you would need to do is find what the common value is that satisfies the constraint, right?

@qq-rf6hv 3 жыл бұрын

Yes

@onkarsingh8358 3 жыл бұрын

Luv u grom India👈

@leopoldschlemmer3387 3 жыл бұрын

Can’t you just state that if a>=b,c then 1 >= 3*a/(1+a), as a/(1+a) is decreasing when a is increasing. If you solve you get a

@Alliedmaths 3 жыл бұрын

a = b = c can be easily solved by doing because it has similarity

@neilgerace355 3 жыл бұрын

0:10 Three of these things belong together Three of these things are kinda the same But one of these things is not like the others Now it's time to play our game It's time to play our game!

@PinkeySuavo 3 жыл бұрын

idk why/how i ended up here at 02:42 am but quite noice btw i was "going to sleep" 30 minutes ago after watching vsauce video lol

@quickyummy8120 3 жыл бұрын

I found another amazing Olympiad video kzfaq.info/get/bejne/hsCeqNd1vZaWo30.html

@lgooch 2 жыл бұрын

I tried to apply Cauchy Schwarz or AM-GM too early, I refused to multiply all of that shit out.

@30indrayudhdas28 3 жыл бұрын

Can we use rearrangement inequality? I think so🤔

@vivekbhutada3049 3 жыл бұрын

Wanted to say you that dont use indian flag vertically in the thumbnail. A humble request !

@kumaracademy7555 3 жыл бұрын

❤️❤️🔥🔥Always love your content

@erikawimmer7908 3 жыл бұрын

Me: doing math. Papa: uploads Video Me: watches math Video abd doing math. ULTIMATE POWER

@1987Videolover 2 жыл бұрын

When i see the question.. I know it will be using AM GM..

@juijani4445 3 жыл бұрын

papa whats the point of including mathematical olympiad in the title? isnt MO in RMO nuff? is this for algorithm??

@PapaFlammy69 3 жыл бұрын

dunno

@baerlauchstal 3 жыл бұрын

It's not *that* crazy-hard to extremise the product of p/(1-p), q/(1-q) and (1-p-q)/(p+q) over the triangular domain p>=0, q>=0, p+q

@rajnikantsharma4054 3 жыл бұрын

What do you think about india? Don't know its power. I will show you the power of indian competitions and competitors both!

@SK-qc2hb 3 жыл бұрын

Great video, but just one thing -- if you perchance show the solution to math problems in JEE, and mention how nice it was, you might face problems, like Jack Fraser did. He didn't say anything against Indians, but it went straight downhill for him for months after his Quora answer.. Nice video!!

@turtlelink6845 3 жыл бұрын

Can also be done using Jensen :) (Should be possible once you sub x = a/(1+a) etc.)

@paulm5441 3 жыл бұрын

what is a Jensen? :)

@raphaelteng2344 2 жыл бұрын

I would just like to start off by saying that among Algebra, Geometry, Number Theory and Combinatorics (Math Oly topics), Algebra scares me the most because the solutions are the shortest and induce the highest intensity of bruh upon seeing the solutions like this one. Basically, I first obtained 1/(1+a) + 1/(1+b) + 1/(1+c) = 2 ---- (1) Then I got the equation you got at 5:40 and did not divide by 4 where is the motivation for that bruhhh this is exactly what I mean I divided the entire equation by abc to get 1 + (1/a + 1/b + 1/c)/2 = 1/2abc ---- (2) Note that 1/a - 1/(1+a) = (1/a)x(1/(1+a)). This gave me the motivation to use Chebyshev's inequality: (1/a)(1/(1+a)) + (1/b)(1/(1+b)) + (1/c)(1/(1+c)) >= ((1/a + 1/b + 1/c)/3)(1/(1+c) + 1/(1+b) + 1/(1+c)) Hence, 1/a + 1/b + 1/c >= 6 ---- (3) after the application of (1) Subbing this into (2), it is trivial that abc

@satyarth7353 3 жыл бұрын

I'm an indian and I can confirm that indian competitive exams are overrated.

@akshat9282 3 жыл бұрын

6 comments. 5 from indians. one from papa flammy ahahah legend

@GammaFZ 3 жыл бұрын

discord gang

@larakalish881 2 жыл бұрын

Ich mag der Inhalt ihnen Videos! Gehen sie weiter! Support from the US! Unterstützung aus Amerika!

@mihaipuiu6231 2 жыл бұрын

I tell you my opinion...you are a smart teacher, ...your write is nice and very visible. You amazed me with your humor, which is a plus . Thanks.

@duggydo 3 жыл бұрын

Since a,b,c are all reals, I just solved with compass and straightedge. Ezpz weeny squeezy!

@rvure 3 жыл бұрын

Do more RMO probs my man

@Djerszium 3 жыл бұрын

Haven't watched the vid yet but 0,1,1 is it I think Cause 0+1/2+1/2=1

@brandonfoy9207 3 жыл бұрын

A mean problem? how can that be hard just calculate the average

@PapaFlammy69 3 жыл бұрын

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@gigagrzybiarz 3 жыл бұрын

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@patricius6378 3 жыл бұрын

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@piyushpathak5012 3 жыл бұрын

As its kinda based on the AM-GM inequality.

@stargazer137 8 ай бұрын

LMAO

@physicslover1950 3 жыл бұрын

Omg that concept of arithmetic mean and geometric mean was great. I think I need to revise everything 😌😌😌😌

@nicholasthesilly 3 жыл бұрын

Look up harmonic means. And their relation to the other two kinds :)

@akinit2978 3 жыл бұрын

I did this one a bit differently: let A = a + 1, B = b + 1, C = c + 1 and then the original equation takes the form: (A - 1)/A + (B - 1)/B + (C - 1)/C = 1 from which we get 3 - 1/A - 1/B - 1/C = 1 2 = 1/A + 1/B + 1/C Now since 2 = 2 * 1, multiply the original by 2 and get (2A - 2)/A + (2B - 2)/B + (2C - 2)/C = 2 (2A - 2)/A + (2B - 2)/B + (2C - 2)/C = 1/A + 1/B + 1/C and from here it follows that 2A - 2 = 1, 2B - 2 = 1, 2C - 2 = 1 => A = 3/2, B = 3/2, C = 3/2 => a + 1 = 3/2, b + 1 = 3/2, c + 1 = 3/2 => a = b = c = 1/2 => abc

@akinit2978 3 жыл бұрын

Ok, I AM retarded, I didn't show that abc

@shayangfkk7948 3 жыл бұрын

Always is the indian guy doing german mathematics . Now german boi , doing indinans .

@scienceshed5466 3 жыл бұрын

7:32 papa spitting the truth 😂 and i can feel that because i am one of child rhat hsa to give those fucking stupid exams that are just 😑

@PapaFlammy69 3 жыл бұрын

@kennytheripper2526 2 жыл бұрын

This is the reason why the Topper of my city couldn't qualify RMO

@srirampraveenva 3 жыл бұрын

Should've just done drugs like a normal teenager instead of studying for these exams smh

@tszhanglau5747 3 жыл бұрын

Can you talk about am-gm inequality in detail?

@PapaFlammy69 3 жыл бұрын

Can do! :)

@patricius6378 3 жыл бұрын

@@PapaFlammy69 and maybe motivate the geometric mean a little? :) bc I've never heard about this thing :O

@jamiejoseph7823 3 жыл бұрын

Lmao the gate for the Indian math tho

@tobiasgorgen7592 3 жыл бұрын

6:28 its the arithmetic meme

@jackhandma1011 3 жыл бұрын

Now prove AM-GM without calculus and induction (idk how to).

@shabushabu1453 3 жыл бұрын

Wait, what? How is (1/4)^4 = (1/2)^2

@drabart6121 2 жыл бұрын

Em, could someone explain why there are 3 lines in the 'equals' sign and he just ignores it? From what I've learnt that means congruence and not equality. (Maybe they use 3 lines in India or sth)

@PapaFlammy69 2 жыл бұрын

identically equal