What we managed to prove is that for any a1, a2, …, a(2^k), the inequality holds This means, when we choose the last term to be the mean of the previous terms, the inequality would still hold, and we can simplify it to the case of a1, a2, …, a((2^k)-1) So you could treat it as a special case of a1 to a(2^k), where it’s only dependent on the first (2^k)-1 terms and turns out to be true
best proof i've seen so far, well done!
Why is 7 afraid of 8?
Induction.
Why ak=(a1+a2+a3+ak-1)/k-1, i cannot wrap my head around this
Same thing
yuhus
It seems great, however why it is correct to assume a_k equal to the mean of previous terms？I just can not convince myself that is proper.
Ya I want to know the same thing!! Can you help me?
It's because a_1, ..., a_k are arbitrary (nonnegative) values, and for said values we assumed that (a_1 + ... + a_k)/k >= kth root of (a_1*...*a_k).
What we managed to prove is that for any a1, a2, …, a(2^k), the inequality holds
This means, when we choose the last term to be the mean of the previous terms, the inequality would still hold, and we can simplify it to the case of a1, a2, …, a((2^k)-1)
So you could treat it as a special case of a1 to a(2^k), where it’s only dependent on the first (2^k)-1 terms and turns out to be true
excellent
E
Nice video!
Nice