How to show $\frac{c}{n} \leq \log(1+\frac{c}{n-c})$ How to show $\frac{c}{n} \leq \log(1+\frac{c}{n-c})$ for any positive constant $c$ such that $0 < c < n$?
I'm considering the Taylor expansion, but it does not work...
 A: Hint: For all $n-c \le x \le n$, we have $\dfrac{1}{n} \le \dfrac{1}{x}$. Hence, $$\displaystyle\int_{n-c}^{n}\dfrac{1}{n}\,dx \le \int_{n-c}^{n}\dfrac{1}{x}\,dx.$$
A: Hint: With $x = (c/n)/(1 - c/n)$
$$\log(1 + x) = \int_1^{1+x} \frac{dt}{t} \geqslant \frac{x}{1+x}$$
A: Starting with 
 $$ e^t\ge 1+t\qquad\text{for all }t\in\Bbb R$$
(possibly the most useful inequality about the exponential)
we find by plugging in $-\frac cn$ for $t$
$$ e^{-c/n}\ge 1-\frac cn= \frac{n-c}{n}=$$
and then after taking reciprocals (both sides are positive!)
$$e^{c/n} \le \frac n{n-c}=1+\frac c{n-c}$$
Finally, apply logarithm to obtain
$$\frac cn\le \ln\left(1+\frac c{n-c}\right) $$
A: By the mean value theorem, $e^{c/n} = 1 + e^{\alpha}(c/n)$ for some $\alpha\in (0, c/n)$. Since $e^{\alpha} \le e^{c/n}$, then $e^{c/n} \le 1 + e^{c/n}(c/n)$. So $(1 - c/n)e^{c/n} \le 1$, or 
$$e^{c/n} \le \frac{1}{1 - c/n} = \frac{n}{n-c} = 1 + \frac{c}{n-c}$$
Now take logarithms.
A: 
I thought it would be instructive to present a way forward that does not rely on calculus, but rather an elementary inequality.  To that end, we proceed.
I showed in THIS ANSWER using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities
$$\frac{x-1}{x}\le\log(x)\le x-1 \tag 1$$
for $x>0$.

Using $(1)$ with $x=1+\frac{c}{n-c}=\frac{n}{n-c}$, and thus $\frac{x-1}{x}=\frac{c}{n}$,  it is easy to see that for $n>c>0$
$$\bbox[5px,border:2px solid #C0A000]{\frac{c}{n}\le \log\left(1+\frac{c}{n-c}\right)\le \frac{c}{n-c}}$$
And we are done!
