Recurrence? Prove that if P(2) and P(n) are valid, so is P(2n)

EliseLahn

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Apr 11, 2021
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Consider the following preposition P(n):

P(n): If x1,..,xn ≥ 0, then x1*x2*···*xn ≤ ((x1+···+xn)/n)^n

See that P(2) is valid because (x1+x2)^2 −4x1x2 = (x1−x2)^2 ≥ 0.

a) Prove that, if P(n) is valid for n >1, then P(n−1) is valid. (Tip: define xn = (x1+···+xn−1)/(n-1).)
b) Prove that if P(2) and P(n) are valid, then P(2n) is valid.
c) Explain why it implies that P(n) is valid for every n.

I need help with items (b) and (c). I don't get how I can get to P(2n) from P(2) and P(n). English is not my first language, so sorry about that. Hope I could make myself understandable. Thank you.
 
I don't get how I can get to P(2n) from P(2) and P(n).
If I were you, I would write out what P(2) and P(4) are, and look for a relationship between them. I suspect you will be able to prove P(4) using P(2) by breaking the 4 variables into 2 groups of 2; and that will suggest how to relate P(2n) to P(2) and P(n) in general.
 
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