A Mean Theorem! Proving the AM-GM-HM Inequalities Elegantly!
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- เผยแพร่เมื่อ 1 ส.ค. 2024
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Theory and Proofs Playlist: • Theory and Proofs
Today we want to proof the very important analysis result called arithmetic - geometric - harmonic mean inequalities. At first we prove a Lemma, which is going to make the main proof of the theorem really trivial :) Enjoy! =D
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0:00 Intro
3:40 Proof of Lemma 69.420
15:45 Proof of AM-GM
20:15 GM-HM Proof
Get a load of THIS guy would ya?
hello .son
You better stop I’m 16
@@thisguy3572 th-cam.com/video/xlEWxwsuFZ8/w-d-xo.html
@@thisguy3572 same age disagree :)
Put this guy back in papas basement will ya
I have wanted to see a proof of mean inequalities for ages. Thank you Papa!
13:20 you can't assume that the product of x_1 ... x_{k-1} is 1.
Consider your proof applied to some set of numbers like : {5/2 , 4/3 , 3/10, 7/3, 9/5, 5/21}, you can't just pull out 2 and still have your induction hypothesis satisfied.
I see students make the same mistake from time to time too. They really really want to use their induction hypothesis so they just 'assume' they can.
Yes, I was wondering about this too!
I'm wondering why hardly anyone is pointing this out. In the last 1.5 years, proofs on this channel have become drawn out with extra focus about little things such as the commutativity of addition and integral substitutions while large details like this are overlooked.
Does anyone know the correct way of doing the induction step?
@@kevinli7758 Use the hypothesis on the k numbers x_1, ..., x_(k-1), x_k x_(k+1) and the inequality Flammy proved.
So x_1 + ... x_(k-1) + x_(k)x_(k+1) >= k. And since x_k + x_(k+1) - 1 >= x_(k)x_(k+1) from his inequality, we know that x_1 + ... x_(k-1) + (x_(k) + x_(k+1) - 1) >= k. Add one to both sides and we've proved the lemma. Did I miss anything?
"Proof of Lemma 69.420" lol
Glad someone noticed :DDD
XD
At 12:54 you claim P(k) implies x_1 + ... + x_k ≥ k and use this inequality in the proof of P(k+1). However, P(k) really only implies x_1 + ... + x_k ≥ k under the assumption x_1 * ... * x_k = 1, which isn't necessarily true in the setting of P(k+1) where x_1 * ... * x_k * x_{k+1} = 1 (and even under the assumptions of "Case II"). Am I missing something?
I think the correct way is to apply P(k) on the numbers x_1, ..., x_{k-1}, x_k*x_{k+1} which do have product equal to 1 and thus have x_1 + ... + x_{k-1} + x_k * x_{k+1} ≥ k. Bounding the last term by x_k * x_{k+1} ≤ x_k + x_{k+1} - 1 by the assumptions of Case II will lead to x_1 + ... + x_k + x_{k+1} ≥ k +1 as desired.
@@bangryakYeah, that's correct. Nice
@@bangryak Same reaction. Great job
Great to see these more interesting theorems! Learn something new here indeed!
So elegant! It all comes down to choosing the right sequence ai and then everything is derived so smoothly
Here's a more general result. This is one of those useful tricks for IMO people
Consider the p-mean M_p(x_1,x_2,...,x_n)=((x_1^p+x_2^p+...+x_n^p)/n)^1/p
Arithmetic mean is M_1(...)
Geometric M_0(...) (has to be done with a limit as p goes to 0)
Harmonic is M_-1(...)
We get a lot more common means like M_2(...) the "quadratic mean" and M_3(...) the "cubic mean" as well by taking limits we have M_-infinity(...) which is the min of all elements and M_+infinity(...) which is the max of all of them.
The statement is:
If p>q then M_p(x_1,x_2,...x_n) >= M_q(x_1,x_2,...x_n) with equality if ALL the x_i are equal to the same number.
We instantly find that the AM-GM-HM inequalities are true since 1 > 0 > -1.
There's an even more general statement saying that the inequalities are still preserved even after adding weights to each x_i, pretty neat
I’ve seen the power mean inequality i never knew the geometric mean was the 0th power mean. That’s really interesting cause it looks so different from the other power means at least on the surface
When you just realize it's afternoon in Germany and Evening in India...
Guten Tag Sir
Ich bin Hardik
Ich bin sechzehn zahre alt..
Love from India Sir 🇮🇳
Herr PapaFlammy🔥
I want to know ur name sir..
@@hardikjoshi8557 Er heißt Jens
@@hardikjoshi8557 Jens Fehlau
@@keshavb3128 Thanks a lot 👍🙏
Your channel with the meme edits is slowly becoming my fav math channel to binge on !
A little pedantic, but in the string of inequalities written at the beginning, the middle expression isn't necessarily defined in a generic ordered field (e.g. the rationals). We could take the nth power of all sides to make it valid.
Damn, this brings me back. That case distinction is slick. I thought you were going to finish it off a bit differently though: just apply P(k) to the numbers (x_1,...,x_{k-1},x_k*x_{k+1}) then use the CII inequality
x_1+...+x_{k-1}+x_k+x_{k+1} \geqslant x_1+...+x_{k-1}+x_k*x_{k+1}+1\geqslant k+1
I really love your videos. Keep them coming, you're making math fun and enjoyable ^_^
Thx a bunch Andre!!! =)
I love seeing this kind of content :DDDD
What an enthusiasm man, makes your lecture wonderfull.
You can't apply the induction hypothesis to {x_1,x_2, ... x_k-1}, the product of these numbers might not be 1.
This was on our test last week, very interesting indeed!
Diese ist sehr gutscheiße. My favorite of your recent videos for sure :)
i just realized Papa must be making these videos upside down, so gravity is pulling the chalkboard "upwards." We've figured out his secrets!
Excellent lecture.
Sir it was the best ... 👍👍👍 Must say your way of teaching is marvelous
Flammabily nyshhh!! 🔥🔥
What blackboard do you use?
thank you, actually forgot the prove i did it 1 year back when i was in 11th and didn't made any notes at that time.
Really interesting, thank you
:)
What type of board is that ?
I really liked that type but couldn't find in the local store ?
Wonderful!! 👏
dude your induction hypothesis couldn't be applied 'cause of the prod of first k - 1 terms must not be equal to 1
papa flammy please do a video on differential forms and wedge product and stuff like that
The intro meme was literally me 90% of the time during writing my thesis
I have an issue with the proof, namely your inductive step in the lemma. You use the inequality for x_1 ... x_k = 1 implies x_1 + ... + x_k ≥ k, but you can't guarantee this if product is equal to one if x_1 ... x_(k+1) = 1 also. Indeed, that would force x_(k+1) = 1, contradicting your assumption about it being bigger than 1. What you should have done instead is use the n=k case by writing x_k*x_(k+1) as one term. Then you get x_1+...+x_(k-1) + x_k*x_(k+1) ≥ k. Then you can use x_k + x_(k+1) ≥ 1 + x_k*x_(k+1) to finish.
thank you
That's amazing.
GM Inequality = Girl-Man Inequality
What does AM and HM mean then?
Mathematics never lies
AM = Aunty-Men inequality
HM = Horse-Men inequality
@@enkaramessi10 HM inequality is housewife -man inequality
based maths
Papa I got your watch and I love it
It’s so funny
I think to be even more humorous you could make the hands move but don’t like them up with the etch marks? Like when it moves it’s always in between two etch marks so you can’t really be certain
hi, you can proof the abel- plana formula?
hey sir I am preparing for jee main and jee advanced but I am unable to do that exams maths questions can you help me and can you taught me mathmatics for iit I am from india
Peoples listen I started watching hims videos and I feeled strange feeling so I checked and i have put on 20 kgs of lean muscle by watching these videos and I went down to training section zone and now i can bench 400 pounds in kilograms and Im become fucking jacked by watching keep him watching everyone
The channel’s name might be Flammable now, but high quality videos like this one are inspiration for what the name used to be 😉😘🥰😍🤪😜🤓🤤
Integrate from 0 to inf of (1/x!)?
@Flammable Maths we achieved Jens' constant on latest poll of channel "What if"
nice.
is there a proof of RMS > AP ?
Papa! Could you do a video about euler product please?
Wow you really did it!
Sure thing! =D
well... there is a huge problem with this "proof": the assumption made at 10:50 does indeed lose generality (for k>1 there is no need for a single multiplicative inverse, eg 0.2*0.5*10=1).
interestingly enough, this is never used later in the proof, but then there is another problem: around 14:00 it is claimed that x(1)+...+x(k)>=k. which is not true here, because x(1)*...*x(k) is NOT 1. indeed, it is less than 1 since x(k+1)>1.
you seem smart
bro you saved my life
Not every ordered field has (positive) nth roots for all its positive elements, e.g. the rationals. Of course, assuming the field we're working in does, the result follows. Sorry, I'm a born nitpicker. Great video, Papa!
P.S. Just noticed the lemma number! 😂
wait are you saying at 10:49 that wlog you have x_k1 and the rest multiplies to 1? i dont see the latter.
i also stuck at that part
Very ingenious proof. But I am still fan of Cauchy's proof due to its simplicity
Father Flame, could you do something on quadrature methods? I think Gaussian quadrature is especially cool. What do you think?
As an engineering student, My soul yearns for approximations... What if we compromised and you talk about the error term so that it’s technically exact, but still an approximation technique.
You would make Daddy Gauss proud.
If you assume that P(k) is true in your inductive step, then x_1...x_k=1, and that means that the next x_{k+1}=1 shouldn't it?
My favorite proof of the cuadratic mean-aritmetic mean inequality, in a similar way to cauchy-scharwz:
Let x1,...xn be positive. Consider the polynomial P(x) = sum (x - x_i)^2. This defines a cuadratic polynomial which is greater than or equal to 0. Thus B^2 xi = xj
There's a much simple way to prove AM ≥ GM ≥ HM.
Let each letter from A to Z be assigned a number from 1 to 1/26 in order such that A = 1, B = 1/2, ..., Z = 1/26. Then, consider that 1 ≥ 1/7 ≥ 1/8. However, using the mapping, we have that this is equivalent to A ≥ G ≥ H. Now, since M = 1/13 > 0, then through right multiplication we have that AM ≥ GM ≥ HM. Q.E.D.
do the generalized mean inequality
13:05 why is P(k) hold imply Lhs that product is 1.
"O.B.d.A." - I didn't realize how much I actually missed it :-)
PS: got your CNC running?
Très bon !
Une preuve Très intuitive et qui colle au problème contrairement à la preuve utilisant exp(x) >= x+1
I see papa is doing great proofs
There's a legend that the Ghost of Euler haunts Papa Flammy's University.
But why have you assumed that x_1*x_2*...*x_k=1 while talking about P(k+1)? The x_i's in the P(k) statement are different from the x_i's in the P(k+1) statement.
There's a legend that the Ghost of Euler still haunts Papa Flammy's university. His powers have gotten stronger! 😨😨😱😱😬😬
I think it’s not possible to use inequality from assumption because otherwise usage of x1*x2*...*xk = 1 => x1+x2+...+xk >= k implies x1*x2*...*xk*x(k+1) = 1, so x(k+1) is not arbitrary but only 1. (Time: 13:02)
I have another proof:
Suppose x1*x2*...*xk = 1 implies x1+x2+...+xk=k, xi > 0
Show that x1*x2*...*xk*x(k+1) = 1 implies x1+x2+...+xk+x(k+1)>=k+1, xi > 0
a) if x(k+1) = 1, it is obviously true
b) If x(k+1) > 1, at least one of x1, x2, ..., xk < 1
Wlog, we can reorder xi: x(k+1) > 1 and x(k) < 1
c) If x(k+1) < 1, do the same thing as in case b
x1*x2*...*xk*x(k+1) = x1*x2*...*[ xk*x(k+1) ]= 1
it implies U = x1+x2+...+ [ xk*x(k+1)] >= k (*)
U = x1+x2+...+ [xk + x(k+1)] + [ xk*x(k+1)] - [xk + x(k+1)] + 1 - 1 = [ x1+x2+...+xk+x(k+1) - 1] + [ xk*x(k+1) - xk - x(k+1) + 1] = [ x1+x2+...+xk+x(k+1) - 1] {1} + (x(k+1) - 1)(xk - 1) {2}
{2}: use case b: (x(k+1) - 1)(xk - 1) = k
x1+x2+...+xk+x(k+1) = k+1
If you are not agree with me, please reply my comment and I will email you to discuss
One more thing I’ve noticed: 14:17
You can use induction hypothesis in the following way to get clean solution:
x1+...+x(k-1)+xk+x(k+1) >= x1+...+x(k-1)+xk*x(k+1) + 1 >= (apply induction hypothesis to x1, x2, ..., x(k-1), xk*x(k+1)) >= k+1
Please teach for mathematical olympiads
0:17 that intro was so cute lmao
Once I tried to prove AM-GM inequality by taking the gradient of the function (GM -AM) and putting it equal to zero, and then doing the Hessian of the function. Unfortunately, the Hessian was zero, so I ended concluding nothing. If someone has a somewhat similar ideia to prove the AM-GM inequality, please share
what about GM > IM > NM, the ELO inequality
It looks easy but I mostly forgot to apply when apply when problem come
What about the QM
Could anyone recommend me a chalk board for me to do maths on thats decently sized and reasonably priced? (About £30) it would be highly appreciated, thanks.
Neat!
Is like if you have an apple and it's nth root is greater than or equal to all the apples in the garden over the number of trees .
Hmmm .. interesting!
I have a suggestion .. why don't you make some course and upload the videos in the channel ? It will be really awesome to learn some real analysis or whatever from Papa flammy !
Dear Papa Flammy, please help me out with this problem:
Find all the pairs (x,y) out of Z, such that
12x^2 - (x^2)(y^2) + 11y^2 = 223.
Thanks in advance.
Just factor the left hand side of the equation:
12x² - x²y² + 11y² - 12 *11 = 223 - 12*11
(y²-12)(11-x²) = 91
Now consider each divisor of 91 and work by cases. You quickly come to the conclusion that only
y² - 12 = 13
11 - x² = 7
gives solutions for x, y ∈ ℤ
So x = ±2 and y = ±5
@@maxsch.6555 Ooooohhhh my god!!!! tf is wrong with me for not seeing that? Anyways. Thanks for that.
Lol I was just studying the same thing this morning
nth roots in an ordered field...
3:43 why is that lemma 69.420?
Although P(1) was trivial, he should have done P(2), and that would have been easy as (x-1)(1-1/x)>=0 for x>1. Rather than just assume there are samples larger or less than one, this is a conclusion from x_1...x_n=1, if all the x_i's>1, this is violated and likewise for x_i's
In your lemma you assumed that an multiplicativ invers exists. if you multiply 2*3*1/6 = 1.
sir i see your video regularly. sir i have a question ,what would be the taylor expansion of (1-x)^1/x ?
sir please help me
The Taylor expansion centered at what point? "The" Taylor expansion of an expression is nonsense, because "the" Taylor expansion is not unique: it changes according to what the center of expansion is.
This can be generalized with some inventiveness. Consider (x(1), ..., x(n)) an element of (0, +♾)^n, and consider f a continuous bijection (0, +♾) -> U, where U is an open subset of R. This means f is strictly monotonic. The generalized f-mean of (x(1), ..., x(n)), also called the quasi-arithmetic mean with respect to f, also called the Kolmogorov mean with respect to f, denoted (M[f])(x(1), ..., x(n)), is equal to [f^(-1)][A(f[x(1)], ...., f[x(n)])], where A(x(1), ..., x(n)) is the arithmetic mean of (x(1), ..., x(n)). The arithmetic mean is the generalized f-mean where f = Id. So A(x(1), ..., x(n)) = (M[Id])(x(1), ..., x(n)). The geometric mean is generalized f-mean where f = log, and the Harmonic mean is the generalized f-mean where f(x) = 1/x. The root-mean-square, or the Pythagorean mean, is the generalized f-mean where f(x) = x^2, and the power p-means in general are the generalized f-means where f(x) = x^p. Other generalized f-means sometimes used include f = exp, as a smooth approximation of the maximum function, and as the mean of the logarithm semiring, among others.
Given this, you can prove a similar inequality, though it is a little more restricted, provided that you can prove that the two corresponding functions are of different growth order or asymptotic order. It is quite neat.
I prefer the proof using Jensen’s inequality on e to the x
P(1) shows that AM=GM not AM>=GM. In fact P(1) shows the equality of AM, GM, and HM.
For inequality AM>GM it should be shown that for any n>1, P(n) shows AM>GM. To do so, as far as I know we have to show first that
P(2) leads to AM>GM
P(3) also leads to AM>GM
Then it is assumed that for k>3, P(k) leads to AM>GM. You jump to this assumption instead, by passing P(2) and P(3).
To anyone who is confused at 13:20, I will explain it to you guys based on my understanding. (At first, I also got confused too!😅😅😅)
Firstly, he assumed X_1*X_2*...*X_k*X_(k+1) = 1, containing k+1 terms.
Then, let X_k*X_(k+1) be a term, maybe we can call it X'''_k.
So, X_1*X_2*...*X_k*X_(k+1) = 1 = [X_1*X_2*...*X_(k-1)]*(X'''_k), which means we already obtain only k terms, not k+1.
By assumption of "p(k) is TRUE", we get...
X_1+X_2+...+X_(k-1)+(X'''_k) = X_1+X_2+...+X_(k-1)+[X_k*X_(k+1)] >= k.
- The End -
AM and GM are studied in Theory of Inequality and equations... which i hate but im studying 😅😅
There is a fatal mistake here, but proof could still work
LMAO the thumbnail tho xD
π/e=1
This is the most common joke on every math TH-camrs's comment section yet, it manages to trigger me everytime.
@@_.Infinity._ its not joke its theorem. And most of the time its π=e so this is something totally different
I love u Papa.❤️
WOW!
^ just wanna comment immediately
kkkkkk
What a beaut proof
Only the root mean square is missing
It’s possible for an infinite number of positive samples to equal 0 though right? There was a part where you claimed that was always positive.
What are you talking about? We are working with finitely many samples here.
GM inequality = Chess grandmaster inequality
hi
hey :3
with LOSS of 🧬rality
Prove using Mathematical Induction ezpz lol
How mean
Question about the proof of lemma 69.420: aren't you (falsely) assuming that if
x_1 ... x_{n+1} = 1
then
x_1 ... x_{n-1} = 1?
I'm talking about 10:50 approx. You say you're using the induction hypothesis, but the induction doesn't say that all subproducts are one, it says that _if_ a subproduct is one, the sum should be greater than the number of samples.
Hello
Lemma 69.420
Noice
like this
Do we have a shirt saying *don't be a d³x/dt³*
Yes, I created those originally :) Take a look at my Teespring shop! :)
@@PapaFlammy69 ok :)