$$\frac{x-1}{x} \leq \ln(x) \leq x-1 \forall\ x>0$$

Consider $$f(x)=\ln(x)-\frac{x-1}{x}$$

$$f'(x) = \frac{1}{x} - \frac{1}{x^2} = \frac{x-1}{x^2}$$

Now consider the following two cases:

Case A: $$0 < x \leq 1$$ and Case B: $$1 < x < \infty$$

Then for Case A: $$x-1 \leq 0$$ and hence $$f'(x)\leq 0$$ $$\implies$$ $$f(x)$$ is a non-increasing function in $$(0,1]$$ and hence $$f(x)\geq f(1) \forall x \in (0,1]$$

So for $$x \in (0,1]$$, $$f(x) \geq f(1)$$ $$\implies$$ $$\ln(x)-\frac{x-1}{x} \geq 0$$ $$\implies$$ $$\frac{x-1}{x} \leq \ln(x)$$

Now for Case B:

$$1 < x < \infty$$ $$\implies$$ $$x-1>0$$

Thus, in this region $$f'(x)>0$$ and hence $$f(x)$$ is a strictly increasing function in $$(1,\infty)$$

Thus, for $$1 < x < \infty$$, $$f(x)>f(1)$$ and hence $$\ln(x)-\frac{x-1}{x} > 0$$

Combine results with that in case A for $$f(x)$$ to conclude: $$\frac{x-1}{x} \leq \ln(x)$$

Now define $$f(x)= \ln(x)-x+1$$ $$\implies$$ $$f'(x) = \frac{1}{x}-1 =\frac{1-x}{x}$$ Whole story of Case A and Case B again. But it should be simple to see that for $$0 < x \leq 1$$, $$g'(x)\geq 0$$ and hence $$g(x) \leq g(1)$$ $$\implies$$ $$\ln(x)-x+1 \leq 0$$ And for $$1 < x < \infty$$, $$g'(x)<0$$ and hence $$g(x)<g(1)$$ $$\implies$$ $$\ln(x)-x+1 < 0$$

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Saket Choudhary is a PhD candidate at the University of Southern California

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