My Favorite Proof of the A.M-G.M Inequality
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- Опубликовано: 23 фев 2024
- The 38th Video
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Finally I found a proof for n terms, not just 2 terms.
George Polya was a master in finding elegant proofs . I had the pleasure of meeting him personally on several occasions.
I knew two proofs of the means inequality. Now I know three; and this is the shortest!
Top quality video!
Elegant
Thanks for this video bro 😉
Top ! Bravo . merci de tout cœur pour cette preuve brillante.
0:01
Combinatorics be like
You ain't seen nothing yet 🗿
* combinatorics needs clever visualisation
* trig needs static memory
* geometry requires "good boeks"
In Brasil theres a book that i consider being sacred, it was written by Augusto Morgado, and was supposed to be read by professors who want to teach their students, but anyone who reads that book can understand perfectly combinatory like it is the easiest subject
Well that was fun, not sure if I understood it so fast. But I like it. I have never come across this inequality before but then I don't talk to many people. On first sight it seems to offer so much but in fact delivers so little. Maybe that's for why I didn't know about it. My interest though is in that you have a statement about additions in correspondence with multiplications. To have even one is interesting. Maybe there is much more to be learned? This is pretty fundamental to be honest, and totally un understood by masses of Mathematicians. There in is a connection here worthy of a deep dive, but many do not have the breathing capacity :-) Of course I am new to this: add and divide is as to multiply and root. Not many people think like that these days methinks. The inequality is obvious though because the division by n is the same across all x where as the nth root is proportional to each x. It can only get smaller. I realise now that this is statistics duhhhh. I was thinking in terms of Prime Numbers and their mystery. But, and to the point: Anything that relates Addition to Multiplication in the higher indices of numbers is like Gold Dust. Perhaps this is the first Sovereign?
By far the best proof
Inequalities is really tough
See inequalities of Romanian Mathematical Magzine
if f(x) is increasing but concave down on some interval including all the xi values you can show pretty easily using the 1st order taylor series with remainder that f(sum (pi*xi)) >= sum (pi *f(xi)). calling that first sum "AM" for convenience, express each xi as AM + ei. note all the remainders are negative where ei is nonzero, and sum (pi*ei) = 0.
f(x) = logx completes the original request; f(x) = -1/x proves the HM-GM relationship.
Nice proof
An amazing proof
I have another one which uses set theory
R+^n gives all R+ series which contain n element
Define a set that
for some s belongs to R+^n
Sum of s = SC
and Product of s = x
We will call that set as P
Here s is a seri not number
We clearly see that all members of P is smaller than SC^n
Therefore there is a smallest reel number that for all x belongs to P x smaller or equal to that reel number We will call that reel number as Pmax
Assume that
SC = x1+x2...xn
Pmax = x1x2...xn
xa≠xb
SC=x1+x2...(xa+xb)/2...(xa+xb)/2...xn
So product of them in P However
((xa+xb)/2)²>xaxb
x1x2......(xa+xb)/2...(xa+xb)/2...xn > Pmax
A contribiction therefore
x1=x2=...=xn
After that as easy as pie
If there is place you can't get that is bad of my A level english.
Now that's Math content.
Genial
this is the only method in my mind that prove AM GM inequality without using mathematical induction
You can also prove this very easily using the concavity of the natural log i think
@@arielsasson3097 That's equivalent to the video proof.
Heres a proof without mathematical induction
uses set theory
R+^n gives all R+ series which contain n element
Define a set that
for some s belongs to R+^n
Sum of s = SC
and Product of s = x
We will call that set as P
Here s is a seri ,not number
We clearly see that all members of P is smaller than SC^n
Therefore there is a smallest reel number that for all x belongs to P x smaller or equal to that reel number We will call that reel number as Pmax
Assume that
SC = x1+x2...xn
Pmax = x1x2...xn
xa≠xb
SC=x1+x2...(xa+xb)/2...(xa+xb)/2...xn
So product of them in P However
((xa+xb)/2)²>xaxb
x1x2......(xa+xb)/2...(xa+xb)/2...xn > Pmax
A contribiction therefore
x1=x2=...=xn
After that as easy as pie
If there is place you can't get that is bad of my A level english.
Pls do harmonic mean instead
Bro, please change that voice
Please use your real voice, AI is just not there yet.
@@spaicersoda7165 True. but it is certainly more consistent than mine lol
@@MathemadicaPrinkipia I'll get better with some practice, don't be too hard on yourself.