Proof of the AM-GM Inequality (ILIEKMATHPHYSICS)

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This series mainly references "How to Prove It" by Daniel Velleman Third Edition. This type of exercise would be representative of one in Chapter 3. This exercise, however, is present in Chapter 6 where a more general form of this inequality is proven.
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14 окт 2024

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Комментарии : 17

@renesperb 2 месяца назад

One could also give a geometrical interpretation: draw a semicircle (= Thales circle)over a side of length a+b and a right triangle with height h , with foot at a . Then by Euclid's theorem h = √(a b) and the radius r of the semicircle is 1/2 ( a+b). Clearly , r > = h ,which is the inequality we have.

@張洪鈞 2 месяца назад

Good day🙂 Thank you for the good proof. of 1.5>1 by generalized mean ❤, basically, let a=b>0, (a+b)/2=2(a)/2=a=a^2^0.5

@artieedwards 2 месяца назад

Thank you! I thoroughly enjoyed this presentation! I think I'm going to learn quite a bit here...

@orangeinks6681 2 месяца назад

Oh nice, I did this one a month ago, nice to see you do it!

@Leo-lf7lr 2 месяца назад

What's a Lemma? I understand it in the context but i've never heard of the word.

@alexandreaussems5657 2 месяца назад

It’s a true statement that’s used to proof other theorem/proposition

@TheWeirdGuy-053 2 месяца назад

great explaination 😃😃😃 loved it

@AbelAbraha-d4f 2 месяца назад

great timing i was having class on that

@sarwarmadani6117 2 месяца назад

Great video I love it at least the part which I understand 😢

@sweetjohn5968 2 месяца назад

what? (sqrt(a) - sqrt(b))^2 >= 0?????

@iliekmathphysics 2 месяца назад

it would've been so much easier if i started with that instead 😭😭

@albertohart5334 2 месяца назад

At 1:23 you can simply rearrange and get the answer…

@QuentinCordeau 2 месяца назад

Can you show how please ?

@albertohart5334 2 месяца назад

@@QuentinCordeau (a+b)^2 >= 4ab Sqrt both sides a+b >=2sqrt(ab) (a+b)/2 >= sqrt(ab) There You would get plus/minus from square root but I think you can dismiss minus case easily

@jhacklack Месяц назад

You first need to prove that if a > b, then sqrt(a) > sqrt(b), if you want to be completely rigorous.

@albertohart5334 Месяц назад

@@jhacklack this is trivial since y= sqrt(x) is injective and monotonic

@jhacklack Месяц назад

@@albertohart5334 yes we both know that but the proof of those two statements is at least as complex as proving that A^2 > B^2 -> A > B Despite appearances, the proof given in the video is as short as a rigorous proof can be.