Prove the AM-GM Inequality: (a1 + a2 + ... + an)/n ≥ (a1a2...an)^1/n First Proof [ILIEKMATHPHYSICS]

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This video is part of the "Intro to Higher Math" series I am making; it mainly references Daniel Velleman's book "How to Prove it" Third Edition. The exercise in this video is given by Exercise 8 of Section 6.2. The way the proof is done in this video is essentially the way it is guided in the exercise.

Also, every time we mention "list of length N" in the video, we mean "list of positive real numbers of length N".

There are many proofs on this; I am planning on making a video in the future giving another proof.

Thanks and enjoy the video!

ILIEKMATHPHYSICS

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3 things:

This is an extremely clever proof, I enjoyed seeing it come together.

The way the 2 added in post disappeared behind your head at 8:50 was super clean.

When talking about the arithmetic mean, 'arithmetic' is an adjective and is pronounced with major stress on 'met' and minor on 'ar'.

benjaminschmutter

Super neat contradiction proof - never seen this before !

skmaths-help

Amazing! I've been trying to find a good explanation of this proof!

orangeinks

The first written proof of the AM-GM inequality was by *Augustin-Louis Cauchy* (1789-1857), in his book _Cours d'analyse._ In his book, Cauchy proved the AM- GM inequality by using forward-backward induction (also known as Cauchy Induction).

davidbrisbane

Great video, once again. Keep it up 💪🏼

davidemasi__

what a nice proof, did you find it by yourself ?

lorenzosaudito

There is an elegant and short proof using Jensen's inequality, which states that if the function f(x) is concave, i.e.
f''(x) ≤ 0 and f(x)≥ 0, f'(x) ≥ 0 then f(sum (ak *wk)) ≤ sum(f(ak)*wk ) . Here sum wk = 1 and wk > 0. We now take f(x) = ln x and
wk = 1/n . Thus, 1/n*ln(sum(ak)) ≤ sum{1/n*ak), which implies (product(ak))^(1/n) ≤ 1/n*sum(ak), which is the desired inequality.

renesperb

I think n has to be > 1...because if a_1=1/4 and n =1 we don't get 1/4 >= sqrt(1/4)...Just saying....correction: the inequality holds at n=1. No problem.

SigmaChuck

how do you know which one is the contradiction?
Edit: Because the main theorem has been proven for 2, 4, 8, 16, ...
and the other opposite-theorem-base-case hase been assumed for a n in between?

gnostic

Prove the AM-GM Inequality: (a1 + a2 + ... + an)/n ≥ (a1a2...an)^1/n First Proof [ILIEKMATHPHYSICS]

Prove the AM-GM Inequality: (a1 + a2 + ... + an)/n ≥ (a1a2...an)^1/n First Proof [ILIEKMATHPHYSICS]...

Prove the AM-GM Inequality: (a1 + a2 + ... + an)/n ≥ (a1a2...an)^1/n Third Proof [ILIEKMATHPHYSICS]...

Prove the AM-GM Inequality (a1 + a2 + ... + an)/n ≥ (a1a2...an)^1/n Second Proof

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