# Guess of an inequation

• August 5th 2009, 07:51 PM
Xingyuan
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.
• August 5th 2009, 10:45 PM
red_dog
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)