Prove $\left(\frac{2}{5}\right)^{\frac{2}{5}}<\ln{2}$ Inadvertently, I find this interesting inequality. But this problem have nice solution?
prove that
$$\ln{2}>(\dfrac{2}{5})^{\frac{2}{5}}$$
This problem have nice solution? Thank you.
ago,I find this
$$\ln{2}<\left(\dfrac{1}{2}\right)^{\frac{1}{2}}=\dfrac{\sqrt{2}}{2}$$
following is  my some nice methods,
use this inequality
$$\dfrac{x-y}{\ln{x}-\ln{y}}>\sqrt{xy},x>y$$
then we let 
$x=2,y=1$
so $$\ln{2}<\dfrac{\sqrt{2}}{2}$$
solution 2:
since
$$\dfrac{1}{n+1}\le\dfrac{1}{2}\cdot\dfrac{3}{4}\cdots\dfrac{2n-1}{2n}$$
then
$$\ln{2}=\sum_{n=0}^{\infty}\dfrac{1}{(n+1)2^{n+1}}<\sum_{n=0}^{\infty}\dfrac{(2n)!}{(n!)^22^{3n+1}}=\dfrac{1}{\sqrt{2}}$$
solution 3
since
$$(1+\sqrt{2})^2(t+1)-(t+1+\sqrt{2})^2=t(1-t)>0$$
so
$$\ln{2}=\int_{0}^{1}\dfrac{1}{t+1}dt<\int_{0}^{1}\left(\dfrac{1+\sqrt{2}}{t+1+\sqrt{2}}\right)^2dt=\dfrac{\sqrt{2}}{2}$$
solution 4:
$$\ln{2}=\dfrac{3}{4}-\dfrac{1}{4}\sum_{n=1}^{\infty}\dfrac{1}{n(n+1)(2n+1)}<\dfrac{3}{4}-\dfrac{1}{4}\left(\dfrac{1}{1\times 2\times 3}-\dfrac{1}{2\times 3\times 5}\right)=\dfrac{7}{10}<\dfrac{\sqrt{2}}{2}$$
solution 5
$$\dfrac{1}{\sqrt{2}}-\ln{2}=\sum_{n=1}^{\infty}\dfrac{\sqrt{2}}{(4n^2-1)(17+2\sqrt{2})^n}>0$$
But $$\ln{2}>\left(\dfrac{2}{5}\right)^{\frac{2}{5}}$$  I can't have this nice solution
Thank you everyone can help.

The Newton-Raphson method for $x=\left(\frac{2}{5}\right)^{2/5}$ is $x_{n+1}=\frac45x_n+\frac{4}{125x_n^4}$. Take $x_0=1$ and assume that each $x_n\geq\left(\frac{2}{5}\right)^{2/5}$. Compute the iterations up to $x_5=0.693145$, using a sufficient level of precision, around $6$ decimals. Finally, compute the upper bound of $e^{x_5}$ with the first $8$ terms of its Taylor series plus another $8$th term.
$$1.999997=\sum_{n=0}^8\frac{(x_5)^n}{n!}+\frac{(x_5)^8}{8!}\geq e^{x_5}\geq e^{\left[\left(\frac25\right)^{2/5}\right]}$$
Therefore, $\ln(2)\geq\left(\frac25\right)^{2/5}$. The exercise is a question of how few arithmetic operations you need to reach the inequality. This method has a total of  $36$ multiplications, $16$ divisions and $14$ additions *, which is feasible to do within an hour or so, despite being rather tedious. 

* Each Newton-Raphson iteration takes $5$ multiplications, $2$ divisions and $1$ addition. For $n>0$, each term in the series takes at most $2$ multiplications, assuming the previous power of $x_5$ and factorial were stored. Each term also requires $1$ division and $1$ addition. The $n=0$ and repeated $n=8$ terms of the series are single additions.
