How to show that $ \ln( \frac{x+1}{x}) * (x+1) - 1$ is non negative and monotonically decreasing we are given an excercise where we have to show that:
$$ \ln( \frac{x+1}{x}) \cdot (x+1) - 1$$
is non-negativ and monotonically decreasing with $x\geq 0$. I took the derivative which yields: 
$$ \frac{x \cdot \ln(\frac{x+1}{x})-1}{x}$$
which didn't tell me a lot. 
What am I missing?
 A: First let $$f(x) = (x + 1) \ln\left(\frac{x + 1}{x}\right) - 1 = (x + 1) \ln\left(1 + \frac{1}{x}\right) - 1$$
Now let us construct some useful identities. Let $t \geq 0$. From the definition of the exponential function:
$$e^t = \sum_{i = 0}^{\infty} \frac{t^i}{i!} \implies e^t - 1 - t = \sum_{i = 2}^{\infty} \frac{t^i}{i!} \geq 0 \implies e^t \geq 1 + t \implies t \geq \ln(1 + t) \, \, \, \, (1)$$
Also note:
$$t \geq \ln(t + 1) \implies \frac{1}{t} - 1 \geq \ln(t^{-1}) \implies 1 - \frac{1}{t} \leq \ln(t) \, \, \, \, \forall \, \, \, \, t \geq 0 \, \, \, \, (2)$$
We can now construct a lower bound for $f(x)$ based $(2)$:
$$\ln(1 + \frac{1}{x}) \geq 1 - \frac{1}{1 +  \frac{1}{x}} =  \frac{1}{x + 1}$$
$$(x + 1)\ln(1 + \frac{1}{x}) \geq  1$$
$$f(x) \geq  0$$
Taking a derivative of $f(x)$ gives us
$$f'(x) = \ln\left(1 + \frac{1}{x}\right) - \frac{1}{x} < 0$$
where the inequality is established by invoking $(1)$.
Therefore $f(x)$ is non-negative and monotonically decreasing for all $x \geq 0$.
