Can you make another video and explain the general case a bit more in detail please. I could follow your reasoning till the general case. I really want to understand this. Thank you. Why the last term wasn't raised to (1/2^a) for instant?! Why you took n< 2^a in the first place, because to me it won't make it the general case but for the number of elements to be powers of two!
We showed AmGm for all powers of two. Now the general case is that we take an n that's not a power of 2 and let is be less than some power of two. Now we take the thing we knew for powers of two that holds true and show that if we plug in a couple of identical values we'll get the inequality for n. That's the idea
![[AM-GM Inequality for n terms] Forwards Backwards Induction - Proof of AM-GM Inequality for n terms](/img/1.gif)
The famous Cauchy proof. Nice!
This guy is the Hans Neimann of Maths
lol
Liked the how you "filled in the gaps" between powers of 2 informally, before you got to the formality of the induction proof.
great proof, i think you forgot to set X to min({a}_{n+1}^{2^\alpha}) to ensure boundedness tho
You're a beast. Thanks!
My favorite proof uses Jensen's inequality (applying to concavity of ln x)
Oh that's actually pretty cool
wow the explanation is making alot of sense there is no blank spaces while following up
can you help?...when should we use iff symbol in our maths equation or proof etc
13:00 nice technique 😃
thanks，this video helps me a lot❤
Can you make another video and explain the general case a bit more in detail please. I could follow your reasoning till the general case. I really want to understand this. Thank you. Why the last term wasn't raised to (1/2^a) for instant?! Why you took n< 2^a in the first place, because to me it won't make it the general case but for the number of elements to be powers of two!
We showed AmGm for all powers of two. Now the general case is that we take an n that's not a power of 2 and let is be less than some power of two. Now we take the thing we knew for powers of two that holds true and show that if we plug in a couple of identical values we'll get the inequality for n. That's the idea
@@ShefsofProblemSolving I see, thank you for the time you took to explain it more clearly.
omggg i love you !!
I never heard of em then I suddenly had to prove it 😂