Proving this inequality without calculus [closed]

Question

For all $x\in(0,1)$, prove that $\ln x<x$.

My attempt:

First step: Assume a function $f(x)=\ln x-x$. Analyzing tells me the function is continuous and non-differentiable at $x=0$.

Second step: $F'(x) =\frac{1}{x}-1=\frac{1-x}{x}.$

This gives me $x=1$ as an extremum. From further analysis, this appears to be a global maximum.

So combining all the info, $x=1$ is the global maximum with the function decreasing at points less than it. That would mean $f(x)$ is decreasing from $(0,1]$ and $(0,1)$. Hence, this proves the inequality.

However, is there a quick non-calculus way to solve this? I thought of using Taylor series but it's not working.

Answer 1

For $x\in (0,1)$, $\log x$ is negative (because $e^t\geq 1$ for $t\geq 0$) and so $\log x<0<x$ follows. (log is base $e$, as always).

Answer 2

Answer to the original problem

For all $x\in(0, 1)$ prove that $\ln(1+x) < x$ without calculus

We have

$$\ln (1+x)<x\iff 1+x<e^x$$

and by $x=\frac1y$ with $y>1$

$$1+x<e^x\iff 1+\frac1y<e^{1/y}\iff \left(1+\frac1y\right)^y<e$$

which is true. Refer to that proof.