<s>[INST] 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.
 [/INST] Using the Taylor series
$$\ln \frac{1+x}{1-x} = \ln (1 + x) - \ln(1 - x) = 2\sum_{k=0}^\infty \frac{x^{2k+1}}{2k+1}, \quad -1 < x < 1;$$
letting $x = 1/3$, we have
$$\ln 2 = 2\sum_{k=0}^\infty \frac{1}{3^{2k+1}(2k+1)}. \tag{1}$$
Using Bernoulli inequality, we have, for all $a > 0$,
$$(2/5)^{2/5} = \frac{1}{a^2}\left(\frac{2}{5}a^5\right)^{2/5}
\le \frac{1}{a^2}\left(1 + \left(\frac{2}{5}a^5 - 1\right)\cdot \frac25\right) = \frac{4}{25}a^3 + \frac{3}{5a^2}$$
with equality if and only if $a^5 = \frac52$.
It suffices to prove that, there exists $a > 0$ such that
$$\frac{4}{25}a^3 + \frac{3}{5a^2} < \ln 2.$$
When $a^5 \approx \frac 52$, from $\frac{4}{25}a^3 + \frac{3}{5a^2} = \frac{4a^5}{25a^2} + \frac{3}{5a^2} \approx \frac{4\cdot (5/2)}{25a^2} + \frac{3}{5a^2} \approx \ln 2$, we have $a^2 \approx \frac{1}{\ln 2} \approx \frac{81}{56} = 1.4464$
and $a \approx 1.2$ where we have used (1) to get $\ln 2 \approx 2\sum_{k=0}^1 \frac{1}{3^{2k+1}(2k+1)} = \frac{56}{81}$. Letting $a = \frac{6}{5}$, it suffices to prove that
$$\frac{4}{25}(6/5)^3 + \frac{3}{5(6/5)^2} < \ln 2.$$
Using (1), we have
$$\ln 2 > 2\sum_{k=0}^5 \frac{1}{3^{2k+1}(2k+1)} = \frac{15757912}{22733865}.$$
It suffices to prove that
$$\frac{4}{25}(6/5)^3 + \frac{3}{5(6/5)^2} < \frac{15757912}{22733865}$$
which is true.
We are done.
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