AOPS一则美妙不等式

来自RainbowNeos。

Given $x>0$, show that$\left(1+\frac{1}{x}\right)^x \left(2+\frac{1}{x}\right) + e >\frac{3(x+1)^x}{x^{x+1}\ln \left(1+\frac{1}{x}\right)}$

证明：
Let $t=1+\frac1x>1$, we need to prove
$$1+t+\frac{e}{t^{1/(t-1)}}\frac{3(t-1)}{\ln t}.$$

And we use AM/GM/Logarithmic Mean/Identric Mean, which

$$A=\frac{1+t}{2},\qquad
G=\sqrt t,\qquad
L=\frac{t-1}{\ln t},\qquad
I=\frac1e,t^{t/(t-1)}.$$

Hence,

$$
1+t+\frac{e}{t^{1/(t-1)}}
=2A+\frac{G^2}{I}
\overset{\displaystyle I<A,\
\frac AG+\frac GI\ge2\sqrt{\frac AI}>2}{>}
A+2G
\overset{\displaystyle L<\frac{A+2G}{3}}{>}
3L
=\frac{3(t-1)}{\ln t}
$$

PS:让我想起了曾经提过的对数平均的一个上界。