Intuition behind logarithm inequality: $1 - \frac1x \leq \log x \leq x-1$

Take the upper bound: $$ \ln {x} \leq x-1 $$ Apply it to $1/x$: $$ \ln \frac{1}{x} \leq \frac{1}{x} - 1 $$ This is the same as $$ \ln x \geq 1 - \frac{1}{x}. $$

If you don't already know that $\log(x)=\int_1^x\frac1t\,dt$, one way to define the logarithm function is: $$\text{$\log(x)$ is the area under the curve $y=\frac1t$ from $t=1$ to $t=x$.}$$ The picture below shows that this area is sandwiched between two rectangles, each of width $x-1$. The smaller rectangle has height $1/x$, while the larger one has height $1$. In other words, we have the following inequalities:

$$(x-1)\cdot \frac1x\le\log(x)\le(x-1)\cdot 1$$

Starting from the fairly well-known, $$1 - y \leq e^{-y}$$ Rearranging, $$1 - e^{-y} \leq y$$ Substituting $y = \ln x$, $$1 - \frac{1}{x} \leq \ln x$$

TADA!