<s>[INST] Derivation of natural log inequality While looking at a proof of this inequality
$(1+\frac{1}{x})^x < e < (1+\frac{1}{x})^{(x+1)}$
The authors take the natural log of on both sides and get
$xln(1+\frac{1}{x}) < 1 < (x+1)ln(1+\frac{1}{x})$
But then they write this last inequality as follows (which I don't understand)
$\frac{1}{1+x} <ln(1+\frac{1}{x}) < 1/x $
How can you re-write the second inequality as the third inequality?
 [/INST] Let's start from the known equation
$$
x \ln\left( 1+ \frac{1}{x} \right) < 1 < (x+1) \ln\left( 1+ \frac{1}{x} \right)
$$
This is, in fact, two different results:
$$
x \ln\left( 1+ \frac{1}{x} \right) < 1 \qquad \text{and} \qquad 1 < (x+1) \ln\left( 1+ \frac{1}{x} \right)
$$
When you divide both sides of the first equation by $x > 0$ and both sides of the second equation by $(x+1)$, you get
$$
\ln\left( 1+ \frac{1}{x} \right) < \frac{1}{x} \qquad \text{and} \qquad \frac{1}{x+1} <\ln\left( 1+ \frac{1}{x} \right)
$$
Can you see it now?
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