Is there a proof that $\ln(3)>1$ that doesn't use the fact $3>e$ or that $e$ is irrational?

Question

I tried to do this using the $y=\frac{1}{x}$ hyperbola, didn't get anywhere though, I did find other things though like $\ln(2)>\frac{1}{2}$ and $\ln(3)-\ln(2)<\frac{1}{2}$.

Answer 1

Applying the Hermite-Hadamard inequality (see also here) to the strictly convex function $x \mapsto 1/x$ gives the lower bound $$ \ln (3) = \int_1^3 \frac{dx}{x} > \frac{2}{(1+3)/2} = 1 \, . $$

This inequality simply uses the fact that the graph of a convex function lies above any of its tangents, here the tangent at the point $(2, 1/2)$:

Answer 2

Yes, just divide the axis between 1 and 3 into 2n intervals of width 1/n. Since 1/t is decreasing, you get $1/t \geq 1/(1+k/n)$ on the interval $[1+(k-1)/n, 1+k/n]$ and thus $$ \ln(3) = \int_1^3 {dt \over t} \geq {1 \over n}\sum_{k=1}^{2n} {1\over 1+{k\over n}} = \sum_{k=1}^{2n} {1\over n+k}. $$ For n = 4, the right sum is already bigger than 1. It is equal to 28721/27720.

Answer 3

Use the Taylor series for $\ln(x)$ at $x=1$:

$$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}\mp...$$

The radius of convergence is $|x|<1$, so $x=2$ won't do. However, take $x=-\frac{2}{3}$, then

$$\ln\left(\frac{1}{3}\right)=-\frac{2}{3}-\frac{2}{9}-\frac{8}{81}-\frac{4}{81}-...<-1$$

where the first four terms are enough to dip below $-1$. Therefore, $\ln(3)=-\ln\left(\frac{1}{3}\right)>1$.

If you are starting from the integral definition for the natural logarithm, you can of course derive the Taylor series around $1$ by integrating the geometric series $\frac{1}{1+x}=\sum_{n\ge 0} x^n$.

Answer 4

$$\begin{align} \ln(3) &= \frac{⅔^1}{1} + \frac{⅔^2}{2} + \frac{⅔^3}{3} + \frac{⅔^4}{4} + ⋯\\ &> \frac{⅔^1}{1} + \frac{⅔^2}{2} + \frac{⅔^3}{3} + \frac{⅔^4}{4}\\ &= 1.037037037037⋯\\ &> 1. \end{align}$$

Answer 5

Using the Taylor series of the logarithm we get for $0 < x < 1$ $$ \ln \left( \frac{1+x}{1-x}\right) = \ln(1+x) -\ln(1-x) = 2 \left( x + \frac 13 x^3 + \frac 15 x^5 + \cdots \right) > 2x \, . $$ Setting $x=1/2$ gives $\log(3) > 1$.

Answer 6

We can also reach the proof by simply applying the well-known inequality $e^x≥x+1$ .

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Using the substitution $x\longmapsto -\ln x,\thinspace x>0$ we have :

$$ \begin{align}&\frac 1x≥1-\ln x\\ \iff &\ln x≥1-\frac 1x \color{#c00}{\tag 1}\end{align} $$

Equality occurs iff, when $x=1$ .

Then, we see that putting $x=3$ leads to :

$$ \begin{align}\ln 3>1-\frac 13=\frac 23\end{align} $$

which is too weak .

Therefore, we want to multiply both side of the inequality $\color{#c00}{(1)}$ by $6$ and see if it works for us :

$$ \begin{align}&\ln x^6≥6-\frac 6x,\thinspace x>0\\ \iff &\ln x \overset{x\thinspace\longmapsto\thinspace\sqrt[6]{x}}{≥}6-\frac {6}{\sqrt [6]{x}}\end{align} $$

Thus, it is enough to show that :

$$6-\frac {6}{\sqrt [6]{3}}>1$$

which is equivalent to :

$$ \begin{align}&3>\left(\frac 65\right)^6\\ \iff &5^6>2\cdot 6^5\\ \iff &15625>15552 \thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\tiny{\blacksquare}\end{align} $$