My Favorite Proof of the A.M-G.M Inequality

The 38th Video
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[@a_man80]

Finally I found a proof for n terms, not just 2 terms.

[@renesperb]

George Polya was a master in finding elegant proofs . I had the pleasure of meeting him personally on several occasions.

[@levysarah2954]

Top ! Bravo . merci de tout cœur pour cette preuve brillante.

[@GicaKontraglobalismului]

I knew two proofs of the means inequality. Now I know three; and this is the shortest!

[@Maths_3.1415]

Elegant Thanks for this video bro 😉

[@Maths_3.1415]

0:01 Combinatorics be like You ain't seen nothing yet 🗿

[@yash1152]

* combinatorics needs clever visualisation * trig needs static memory * geometry requires "good boeks"

[@beniocabeleleiraleila5799]

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

[@ericv1686]

Thanks for this easy and luminous proof with n variables, far beyond those terrible and endless proofs using induction... Thanks too for the pictures of George Polya, a mighty and inspired mathematician...

[@Maths_3.1415]

Inequalities is really tough See inequalities of Romanian Mathematical Magzine

[@alphalunamare]

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?

[@小衫-n2q]

this is the only method in my mind that prove AM GM inequality without using mathematical induction

[@arielsasson3097]

You can also prove this very easily using the concavity of the natural log i think

[@El0melette]

@@arielsasson3097 That's equivalent to the video proof.

[@Omer-dv2ef]

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.

[@drynshock1]

By far the best proof

[@goncalofreitas2094]

Top quality video!

[@theupson]

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.

[@gregevgeni1864]

Nice proof

[@samueldeandrade8535]

Now that's Math content.

[@drynshock1]

Bro, please change that voice

[@Omer-dv2ef]

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.

[@motihalpert501]

How can we know, Infinitesimal calculus, does not use A.M-G.M Inequality in calculating the limits and derivatives and theorems on which calculus is based? If so, then we are proving a theorem, by something that relies on it and that is incorrect

[@nicolascamargo8339]

Genial

[@spaicersoda7165]

Please use your real voice, AI is just not there yet.

[@MathemadicaPrinkipia]

@@spaicersoda7165 True. but it is certainly more consistent than mine lol

[@spaicersoda7165]

@@MathemadicaPrinkipia I'll get better with some practice, don't be too hard on yourself.

[@berkaymeral9145]

Pls do harmonic mean instead