if $0 < x <1$, How can I prove $ \ln{x} > 1- \frac{1}{x}$ without derivative or integral

Question

If $0 < x <1$, how can I prove$$ \ln x > 1- \frac{1}{x}$$ without derivative or integral?

In the process of proving $$\sin x ^ {\sin x}> \cos x ^ {\cos x}$$ without calculus, I had to solve the above equation, but the idea does not come to mind.

Answer 1

Hint: Take the upper bound: $$ \ln {x} \leq x-1 $$ Apply it to $1/x$

Or

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

---

For given upper bound

To prove: $lnx \leq x - 1$ for $x > 0$.
$\ln(x) < x−1$ for all $x>1$ can be done by contradiction (not required for your question). At $x = 1$ we have equality, so consider $x \in (0, 1)$. Then $0 < 1 - x < 1$. So, using a power series expansion for $ln(1 - x)$ at $1 - x$ we have: $\ln x = \ln(1 - (1-x)) = -(1-x) - \dfrac{(1-x)^2}{2} - \dfrac{(1-x)^3}{3} - \dfrac{(1-x)^4}{4} -\cdots-\dfrac{(1-x)^n}{n}-\cdots < -(1-x) = x - 1$

Or

$y=x-1$ is the equation of the tangent to the ln curve at $(1,0)$ and the function is concave, hence its graph is under the tangent.

Or

You can use $$\tag1e^x\ge 1+x,$$ which holds for all $x\in\mathbb R$

Answer 2

Since every $x\in(0,1)$ can be written as $e^{-t}$ for some $t>0$, your inequality is equivalent to $-t> 1-e^t$ or to $e^t>t+1$, which follows from Bernoulli's inequality (or from the convexity of the exponential function, which can be proved through the midpoint-convexity, i.e. the AM-GM inequality, and the continuity):

$$ e^t > \left(1+\frac{t}{n}\right)^n \geq 1+n\cdot\frac{t}{n} = 1+t.$$