A Mean Theorem! Proving the AM-GM-HM Inequalities Elegantly!

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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

Video sponsored by Brilliant btw :)

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Postfach 11 15
06731 Bitterfeld-Wolfen
Saxony-Anhalt
Germany
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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

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Комментарии

I have wanted to see a proof of mean inequalities for ages. Thank you Papa!

gigagrzybiarz

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.

typha

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?

bangryak

Great to see these more interesting theorems! Learn something new here indeed!

RCSmiths

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

alexismiller

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 k+1
I really love your videos. Keep them coming, you're making math fun and enjoyable ^_^

andreben

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.

steve

What an enthusiasm man, makes your lecture wonderfull.

TheJara

You can't apply the induction hypothesis to {x_1, x_2, ... x_k-1}, the product of these numbers might not be 1.

Nick-kgsk

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.

EssentialsOfMath

So elegant! It all comes down to choosing the right sequence ai and then everything is derived so smoothly

brankoco

Your channel with the meme edits is slowly becoming my fav math channel to binge on !

ahsanhabibkhan

This was on our test last week, very interesting indeed!

xinghuashuying

i just realized Papa must be making these videos upside down, so gravity is pulling the chalkboard "upwards." We've figured out his secrets!

atomic_soup_juice

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 🇮🇳

hardikjoshi