 Juan Hewlett

2021-12-30

Proving an inequality without an integral:
$\frac{1}{x+1}\le \mathrm{ln}\left(1+x\right)-\mathrm{ln}\left(x\right)\le \frac{1}{x}$ ol3i4c5s4hr

Let $f\left(x\right)\phantom{\rule{0.222em}{0ex}}=\mathrm{ln}\left(x\right),$ , then the Mean Value Theorem (Differentiation) says that there exists some $\xi \in \left(x,x+1\right)$
$\mathrm{ln}\left(1+x\right)-\mathrm{ln}\left(x\right)=\frac{\mathrm{ln}\left(1+x\right)-\mathrm{ln}\left(x\right)}{\left(1+x\right)-x}=\frac{f\left(1+x\right)-f\left(x\right)}{\left(1+x\right)-x}=\frac{df}{dx}\xi$
As $\frac{df}{dx}\left(x\right)=\frac{1}{x}$ and as the function $\frac{1}{x}$ is monotonic we know that $\frac{1}{x+1}\le \frac{df}{dx}\left(\xi \right)\le \frac{1}{x}$ jean2098

If we write $t=\frac{1}{x}$ the inequalities become
$\frac{t}{t+1}\le \mathrm{ln}\left(1+t\right)\le t,t>0$
Observe that the three expressions are equal when $t=0$, and then show, by differentiating, that $t-\mathrm{ln}\left(1+t\right)$ and $\mathrm{ln}\left(1+t\right)-\frac{t}{1+t}$ are increasing functions. Vasquez

This is trivial using the definition of $\mathrm{ln}\left(x\right)$ as the area below the hyperbola $\frac{1}{x}$ between 1 and x.
$\mathrm{ln}\left(x+1\right)-\mathrm{ln}\left(x\right)$ is the hyperbola area between $x$ and $x+1$, bounded by the rectangles with area $\frac{1}{\left(x+1\right)}$ and $\frac{1}{x}$

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