Prove that $\ln2<\frac{1}{\sqrt[3]3}$

Solution 1:

$$\log(2)=\int_{0}^{1}\frac{dx}{1+x}\stackrel{\text{Holder}}{<}\sqrt[3]{\int_{0}^{1}\frac{dx}{(1+x)^{9/8}}\int_{0}^{1}\frac{dx}{(1+x)}\int_{0}^{1}\frac{dx}{(1+x)^{7/8}}} $$ leads to a stronger inequality than $\log(2)<3^{-1/3}$.

Solution 2:

Setting $s=\sqrt[3]{3}$, you can try seeing whether $2<e^{1/s}$ by using a suitable truncation of the Taylor series. At degree $5$ we have $$ 2<1+\frac{1}{s}+\frac{1}{2s^2}+\frac{1}{18}+\frac{1}{72s}+\frac{1}{360s^2} $$ that is, $$ 360s^2<360s+180+20s^2+5s+1 $$ or $$ 340s^2-365s-181<0 $$ which is satisfied so long as $$ s<\frac{365+\sqrt{379385}}{680} $$ Now proving that $$ \left(\frac{365+\sqrt{379385}}{680}\right)^3>3 $$ is just (very) tedious computations, but they don't need more than pencil and paper.

Solution 3:

Inequalities like this can obviously be "proved" by plugging numbers into a scientific calculator, which means they can also be established, at least in principle, invoking pretty much any convergent Taylor series for the functions involved, with appropriate error bounds. The challenge is organize things so that the arithmetic stays manageable. Here is one attempt to do so.

It's convenient to begin by noting that

$$\ln2\lt{1\over\sqrt[3]3}\iff3\ln2\lt\sqrt[3]9=2\left(1+{1\over8}\right)^{1/3}$$

To get started, we have

$$\begin{align} \ln\left(1+x\over1-x\right)&=\ln(1+x)-\ln(1-x)\\ &=\left(x-{1\over2}x^2+{1\over3}x^2-\cdots\right)+\left(x+{1\over2}x^2+{1\over3}x^3+\cdots\right)\\ &=2\left(x+{1\over3}x^3+{1\over5}x^5+{1\over7}x^7+\cdots\right)\\ &\le2x+{2\over3}x^3+{2\over5}x^5+{x^7\over3(1-x)} \end{align}$$

(where we've generously changed the $7$ to a $6$ and bounded the remainder with a geometric series). Thus

$$3\ln2=3\ln\left(1+{1\over3}\over1-{1\over3} \right)\le2+{2\over3^3}+{2\over5\cdot3^4}+{1\over2\cdot3^6}=2+{20\cdot3^3+4\cdot3^2+5\over2\cdot5\cdot3^6}\\\lt2+{20\cdot3^3+4\cdot3^2+6\over2\cdot5\cdot3^6}=2+{90+6+1\over5\cdot3^5}=2+{97\over3^2\cdot135}$$

On the other hand

$$(1+x)^{1/3}=1+{1\over3}x-{1\over9}x^2+{5\over81}x^3-\cdots\ge1+{1\over3}x-{1\over9}x^2$$

and thus

$$2\left(1+{1\over8}\right)^{1/3}\ge2+{1\over3\cdot4}-{1\over3^2\cdot32}=2+{3\cdot8-1\over3^2\cdot32}=2+{23\over3^2\cdot32}$$

It follows that $\ln2\lt1/\sqrt[3]3$ if $97/135\lt23/32$. This can be finished off with some straightforward multiplication. But it's easier (or more fun) to check that

$${97\over135}\lt{23\over32}\iff{38\over97}\gt{9\over23}\iff{21\over38}\lt{5\over9}\iff{17\over21}\gt{4\over5}\iff85\gt84$$