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.

[@GicaKontraglobalismului]

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

[@renesperb]

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

[@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"

[@levysarah2954]

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

[@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.

[@Maths_3.1415]

Inequalities is really tough See inequalities of Romanian Mathematical Magzine

[@user-vy9ps2je9u]

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.

[@samueldeandrade8535]

Now that's Math content.

[@nicolascamargo8339]

Genial

[@berkaymeral9145]

Pls do harmonic mean instead