Fold the AM-GM inequality

This is a short, animated visual proof demonstrating the arithmetic mean geometric mean inequality using paper folding.

Thanks!

For other visual proofs of this fact see:

This animation is based on a visual proof by Yukio Kobayashi from the 2002 issue of the Mathematical Gazette (page 293).

#manim #shorts #mathshorts​ #mathvideo​ #math​ #mathematics #amgminequality #mtbos​​ #animation​ #theorem​ #pww​ #proofwithoutwords​ #visualproof​ #proof​ #iteachmath #calculus #inequality

To learn more about animating with manim, check out:

[kaaristotelancien]

I'm rediscovering mathematics thanks to these visual demonstrations; I regret not having seen this during my studies

[RandyKing]

several geo facts in this animation! this will be a good exercise for my geo classes…lots of places to pause and look for opportunities to simplify and predict next steps…thanks!

[Ascg]

I was just discussing am and gm relation with my maths teacher and now i saw this
What a coincidence

[kaihiwatari]

Best channel in the entire world . I just came Across this channel and now I am addicted to this videos .
You sir are doing a fantastic job of explaining mathematical concepts with animation . Hats off to you . Please keep posting more . 🙏

[__christopher__]

What is nice is that you immediately see that the difference is given by a similar triangle with side legth the difference of those square roots. And indeed, if you take half the square of that algebraically, you see that it indeed gives the difference of the means.

[bigsiege]

Quite convenient when setting boundaries for the method of chords.

[tamirerez]

AM is always bigger the GM (or equal)
To the channel owner: you can also show the semicircle prove, when the radius of the semicircle is the AM of a and b, and all other cords are the GM.

Thank you for good video 👍❤

[עמיתלרמן]

Here is another solution which I have already talked about its construction in a previous video. Take an a+b long segment. Expand it into a circle of a+b diameter. Now a steps from the end of the diameter draw a perpendicular. The perpendicular is of length √(ab), an exercise for tou to show. But its obviously less than the radius of the circle, since the radius is the longesy perpendicular line from a diameter to the circle. So √(ab)≤(a+b)/2.

It also makes it clear that the only equality case is a=b since the radius and segment become the same.

[adrian_kaye]

The best part of watching is how I dont know any of the formulas and most of the time I come up with them myself

[edwinov]

Imagine an integer number. Please, do it now. Done? You have just created an imaginary number!

[abdelkarimdebbah]

Can't we come up with a formula to calculate the area of the exceeding triangular part? It's consistent despite the values of a and b isn't it?

[sayedmahdiyartabasihaeri]

But a should not be equal to b

That's really cool overall

[-ShootTheGlass-]

I don’t understand the starting point (triangle) and why select it. I understand AM and GM, but how do you know where to start the proof?