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@ai yo I had what I thought was a pretty funny inequality proof and I wanna know if it works. Long story short, I want to show that for all real numbers greater than 0 x, we have that (1 + x)^(1 + x) > e^x, so I substituted in lim n -> oo ( 1 + 1/n)^n for e, said that x >= lim n -> oo 1/n, and so have ( 1 + x)^(1 + x) > (1 + x)^x >= (1 + 1/n)^(nx)
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@ai @MercurialBuilding (1 + 1/n)^(n+1) is strictly bigger than (1+ 1/n)^n, it increases as n increases.
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@ai @MercurialBuilding I think you made some typos in your last paragraph friend.
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@ai @MercurialBuilding I must rain on this parade.
If MercurialBuilding's argument works, it seems to imply the *stronger* claim that (1 + x)^x >= e^x for all x. This is simply not true, for example if x=1.
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@ceo_of_monoeye_dating @MercurialBuilding Still, it is getting at something... Here’s what I’m thinking: the inequality rewrites as
(1 + x)^(1/x + 1) > e.
Set x = 1/n. This becomes
(1 + 1/n)^(n + 1) > e.
Of course, we know e > (1 + 1/n)^n. I never knew that incrementing the exponent by 1 is always enough to flip the inequality. I wonder if there’s some conceptual way to show that (1 + 1/n)^(n + 1) is decreasing as n increases, analogous to the usual “compound interest” argument for why (1 + 1/n)^n is increasing.
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@MercurialBuilding I am a big fan of the idea, but there is part of the proof which I don't buy (and don't see how to fix):
(1 + x)^x >= (1 + 1/n)^(nx)
Why? I agree that x > 1/n, but you'd also want x >= nx in the exponents, and that's not true because we're taking n "large" and x > 0.
I agree with the overall idea: it suffices to prove that, for every x and every n, we have (1 + x)^(1 + x) > (1 + 1/n)^(nx). But, when x is fixed and n is large, 1/n becomes small but nx becomes large. The difficulty seems to lie in this tradeoff.
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Fwiw, here's a boring proof: First I take logs and rearrange to get to
log(1 + x) > x / (1 + x).
Both sides agree when x = 0, so it suffices to prove that the derivative of the LHS is greater than the derivative of the RHS. This is
1 / (1 + x) > 1 / (1 + x)^2,
which is true because x > 0.
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@ai I'm just unsure on how the x and the n interact, heh
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