#Proving the AMGM inequality by proving forwards
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you've only proven it for the case where c is (a+b)/2 though
like you have not actually proven three-variable am-gm
that's the problem it assumes that c is the mean of a and b
but for the cauchy proof it also assumes that the Xn term is the mean
uh that's substituting a value for xn into the n-variable case and deriving the n-1 variable case from it
but isnt this term the AM of the previous terms
so once again it only proves the case where the final term happens to be the AM of the previous terms
no but the final term doesn't exist
like you're explicitly reducing it down into the n-1 variable case
it's like if I wanna go from
x^2 + y^2 >= 0
to x^2 + 4 >= 0
I'm allowed to do that
but I can't say "set y=2, since x^2+4 >= 4 we have x^2 + y^2 >= 4"
like have you noticed that the variable xn does not exist in the final line/conclusion of the statement
...it literally states that it's assuming the AM-GM inequality holds for n terms. That is, for n arbitrary terms, in particular when the nth term is the arithmetic mean of the other terms. And then proving that the AM-GM inequality holds for n - 1 terms.
alright so say we try to prove 3-way AMGM from 4-way
assume a+b+c+d >= 4(abcd)^(1/4)
then let d = (a+b+c)/3
what happens if d isn't (a+b+c)/3
then you get a different and equally valid case of the 4-way AMGM that is most likely going to be fucking useless for proving the 3-way case
like sure yeah you can do that and it’ll work but that’s entirely not the point
please refer to this analogy if you are in the habit of understanding analogies
ohh wait
so letting d=(a+b+c)/3 is just one specific case of the 4-way that happens to reduce into the 3-way case nicely
and my original proof here assumes one case for a 3-way AMGM that isn't universally true
?
ok nah im just slow sry about that
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