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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.

Dongmin Choi
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    would need a definition of $\ln x$ not involving an integral. [usual integral of $1/t$ from $1$ to $x$ would be out?] – coffeemath Sep 09 '19 at 07:15
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    Your requirement is odd looking at the kind of inequality you have. By the way, the original, trigonometric, inequality is also pretty weird. For example, it tells us that for $;x=-\pi/6;$ we have $$\left(-\frac12\right)^{-1/2}>\left(\frac{\sqrt3}2\right)^{\sqrt3/2}$$ yet the left side is not even defined in the real numbers (in the complex there is no "bigger than" thingy). You must be much clearer with the problem's conditions. – DonAntonio Sep 09 '19 at 07:25

2 Answers2

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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 $x\in\mathbb R$

IamKnull
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    How do you know that is an upper bound? How do you know "the function" (which one?) is concave? How can you know the last inequality? I mean, how can you know all this without differential calculus? This would help the OP... – DonAntonio Sep 09 '19 at 07:41
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    I don't know if you can but I think it'd be ridiculous to use power series or Taylor-Maclaurin series in an answer that requires not to use derivatives... – DonAntonio Sep 09 '19 at 08:07
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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):

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

Jack D'Aurizio
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