TITLE: When $x$ is a real number and $x>1$, why is $x^x>(x+1)^{x-1}$?
QUESTION [8 upvotes]: When $x$ is a real number and $x>1$, why is the following true?
$x^x>(x+1)^{x-1}$
I tried finding the minimum of $x^x-(x+1)^{x-1}$ with my limited calculus knowledge, but it shortly appeared out of my range.
It's good when I can understand a good answer, but I'd still be happy to come back years later when I'm better at math, so please don't hesitate to share your knowledge.

REPLY [2 votes]: Let $f(x)=x\ln x-(x-1)\ln(x+1),\;$ so $\color{red}{f(1)=0}$
and $\displaystyle f^{\prime}(x)=x\left(\frac{1}{x}\right)+\ln x-(x-1)\cdot\frac{1}{x+1}-\ln(x+1)=\frac{2}{x+1}-(\ln(x+1)-\ln x)$.
Since $\displaystyle \ln(x+1)-\ln x<\frac{1}{x}\;\;$ (by considering the area under $y=\frac{1}{x}$ from $x$ to $x+1$), 
$\displaystyle \color{red}{f^{\prime}(x)>\frac{2}{x+1}-\frac{1}{x}=\frac{x-1}{x(x+1)}>0}\;$ for $x>1$;
so  for $x>1,\;\;$ $\color{red}{f(x)>0}\implies x\ln x>(x-1)\ln(x+1)\implies x^x>(x+1)^{x-1}$