# Guess of an inequation

• Aug 5th 2009, 07:51 PM
Xingyuan
Guess of an inequation
For any $\displaystyle n \in N$ and $\displaystyle n \not= 0$,for any $\displaystyle 0 \leq t \leq n$,
can we get the inequation: $\displaystyle (1 - \frac{t}{n})^{n} \leq (1 - \frac{t}{n+1})^{n+1}$ ?

I think the inequation is correct,but i can't find a proof of it.
• Aug 5th 2009, 10:45 PM
red_dog
Let $\displaystyle f(x)=\left(1-\frac{t}{x}\right)^x, \ x>0, \ 0\leq t\leq x$

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

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

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

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

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

So, $\displaystyle f(n)<f(n+1)$