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.

Hmm, first I thought the following would allow to compute the proof mentally, but, well...  although I find it a remarkable simplification I'll need the pocket calculator in the end.  But let's see:     
$$\ln(2) \gt (2 / 5)^{2 /5} = \left({16 \over 100 }\right)^{1/5} $$ We have also $$
\ln(2) = \ln \left( 1+1/3 \over 1-1/3 \right) =  2\left( {1\over3} + {1\over 3^3 \cdot 3} + {1 \over 3^5 \cdot 5}+ \cdots  \right) \\ = {2 \over 3} \left( 1 + { 1 \over 9\cdot3 } + {1 \over 81 \cdot 5}+ \cdots ) \right)  $$
Turn the factor 2/3 to the the rhs then
$$ 1 + 1/27 + 1/81/5+ 1/729/7 \cdots \gt \left( {3^5 \over2 \cdot 100 }\right) ^{1/5}= \left( 1+{43\over 200}\right)^{1/5} $$
Now in general we have for a fifth root
$$ (1+x)^{1/5} = 1 + x/5 - 2(x/5)^2 + 6(x/5)^3 - 21(x/5)^4 + 399/5 (x/5)^5 - \cdots $$ 
Thus we must evaluate 
 $$ 1 + 1/27 + 1/81/5+ 1/729/7 \cdots \gt  1 +43/1000 - 2(43/1000)^2  + \cdots  $$
"In principle" this can be done with paper & pen only because the terms decrease quickly, and some adaptions of denominators are possible, however, that was too tidy for me. I found using a calculator (Pari/GP) that we must evaluate the lhs with 4 terms and the rhs with 5 terms (of course excluding the 1's) to get the decision - because after that the partial sums in the lhs still increase but in the rhs decrease.
