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.

Take $f (x) = \frac {5} {2} \log x + \log 5 - x$. Obviously, $f$ is defined on the interval $]0, \infty[$. Since $f ' (x) = \frac {5} {2x} - 1$, it is easy to see that $f$ increases on $]0, \frac {5} {2}]$, take its maximum at $x = \frac {5} {2}$ and decreases on $[\frac {5} {2}, \infty[$. Also, since $f_{\max} = f (x_{\max}) > 0$ and $f (0+) = f (\infty-) = - \infty$, there are exactly two non-negative real numbers $x_1 < \frac {5} {2}$ and $x_2 > \frac {5} {2}$ such that $f (x_1) = f (x_2) = 0$. Thus, we conclude that $f (x) \geqslant 0$ on $[x_1, x_2]$. Since $x_1 < (2/5)^{2/5} < x_2$, we should have $f \left ((2/5)^{2/5} \right) > 0$, that is, $$\log 2 - (2/5)^{2/5} >0,$$ as desired.
