Arithmetic geometric mean inequality

In this video I give an elementary proof of the arithmetic-geometric mean inequality using calculus 1 techniques. This inequality states that the arithmetic mean of a list of positive numbers is always greater than or equal to its geometric mean.

I would like to thank Alex Zorba for providing me with the idea of the proof!

[azmath]

Hi Dr Peyam. Thank you so much for posting this video & for mentioning me. Great work as always.

[janiruliyanage]

This proof is so beautiful because it is so simplistic and ultimately unexpected :)

[saranyak]

It is such a wonderful proof...Thank you so much Dr Peyam and Alex Zorba for this!

[edwinlin]

Hey Dr. Peyam. I found a little article you wrote back in Berkeley for Math 1A called "So you think you can Slant (Asymptote)?" in 2011 and It's been incredibly useful for helping students in Calculus (I tutor kids). But I noticed that this method for slant asymptotes really isn't in any textbook I've ever seen for calculus, nor have I seen it taught formally. Do you happen to know why? Like is it b/c slant asymptotes are just introduced earlier than limits are and it just never gets brought up again?

[Koisheep]

Dr. Peyam, are you familiar with the short proof for the case n=2? Basically, you use that (p-q)² >= 0, take p=sqrt(a), q=sqrt(b) and expand the left-hand side.

The thing is, I tried doing the same for n=3 and OF COURSE it was a monster that led me nowhere. So when I saw your proof I thought "how is the exponential related to all this?" at first to then turn into awe.

[danielcohen]

you are a great teacher. keep making more please.

[isaiahmrman]

This would be pretty neat for some series.

[BomberTVx]

Isn't there a simpler proof?
Coming up with the fact that e^(x-1) >= x
And then drawing the connection to the mean inequalities seems very forced
Isn't there something with induction maybe?

[rolfdoets]

Also a very nice proof! thanks for sharing.

[jameswilson]

Excellent as always! Thanks for sharing!

[pranavsuren]

It's so cool and weird how we use e here.

[ethancheung]

Thank you for shooting this at good angle. Also the use of this f(x) is great

[MrCigarro]

Thank you for this video. In Statistics this is a very important fact.

[ffggddss]

There's a fun & useful way of thinking of various types of means – namely, transformed means.
Like Dr. Peyam is doing in this vid, let's stick to positive numbers only. The ordinary (arithmetic) mean of a₁...a_n is
AM = [1/n] ∑ᵢ₌₁ⁿ aᵢ
If you have a continuous, monotonic function, f, on the positive reals (into the not-necessarily positive reals), then the f-transformed mean is:
fM = f⁻¹([1/n] ∑₁ⁿ f(aᵢ))
In other words, you take the numbers,  transform them with f, take the ordinary AM, then un-transform that with f⁻¹.

And in fact, the geometric mean is the transformed mean using the logarithm as f; in other words, the logarithmic mean. And the base of that log is irrelevant, as long as it's ≠ 1 (base = 1 would be nonsensical anyway), and > 0. Using natural log,
ln-M = GM = e^( [1/n] ∑₁ⁿ ln(aᵢ) ) = e^( [1/n] ln(∏₁ⁿ aᵢ) ) = (∏₁ⁿ aᵢ)^(1/n)

The harmonic mean is the transformed mean using the reciprocal function, f(x) = 1/x  [or = c/x, where c is any non-0 constant.]
HM = 1/( [1/n] ∑₁ⁿ (1/aᵢ) ) = n / ∑₁ⁿ (1/aᵢ)
[ Note that in this example, f is monotonic _decreasing._ ]

And the root-mean-square is the transformed mean using the square function, f(x) = x²
RMS = √( [1/n] ∑₁ⁿ aᵢ² )

These are just a few of the most common transformed means; feel free to play around with others!

PS. I think this property of the GM, along with the fact that ln(x) is convex downward, might also be used to prove Dr. Peyam's proposition.
Anyone see how to do that?

[TALKmd]

Amazing proof, actually its also in part because of your passionate explanation !

[leonardromano]

I think it can also be proven pretty neatly with the fact that ln(x) is a concave function and it's group homorphism property between multiplicative and additive groups. I think I had to prove it in my calc I class and believe I did it that way, but it's already a long time so I don't really remember.

[ybagga]

Can you please explain other proofs as well.. (not one Cauchy induction)

[pedrocusinato]

Make more videos about fractional derivatives pls. I can't find anything about this and it's a really cool thing

[mmukulkhedekar]

amazing!!! Can you do a video and explain rearrangement inequality

[bdcookie]

ur amazing!! with ur proofs i can finish my homework way faster XD