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Math Help - Guess of an inequation

  1. #1
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    Post Guess of an inequation

    For any n \in N and n \not= 0,for any 0 \leq t \leq n,
    can we get the inequation: (1 - \frac{t}{n})^{n} \leq (1 - \frac{t}{n+1})^{n+1} ?

    I think the inequation is correct,but i can't find a proof of it.
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  2. #2
    MHF Contributor red_dog's Avatar
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    Let f(x)=\left(1-\frac{t}{x}\right)^x, \ x>0, \ 0\leq t\leq x

    f'(x)=\left(1-\frac{t}{x}\right)^{x-1}\left[\frac{t}{x}+\left(1-\frac{t}{x}\right)\ln\left(1-\frac{t}{x}\right)\right]

    \left(1-\frac{t}{x}\right)^{x-1}>0 (1)

    Let \frac{t}{x}=y, \ 0\leq y\leq 1 and g(y)=y+(1-y)\ln(1-y)

    g'(y)=-\ln(1-y)>0\Rightarrow g is increasing \Rightarrow g(y)>g(0)=0\Rightarrow \frac{t}{x}+\left(1-\frac{t}{x}\right)\ln\left(1-\frac{t}{x}\right)>0

    From (1) and (2), f'(x)>0, therefore f is increasing.

    So, f(n)<f(n+1)
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