Does this $ n^n >(n+1)^{n-1}$ hold for any real n, including negative and non-natural (rational, irrational)?
The case when $ n \geq 2$ is proved here.
My guess is that it is true only for all natural $n>=2$.
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1No it doesn't because $n^n$ is not even defined for, say, $n=-\frac12$. The question is, however, an interesting one if you restrict $n$ to be strictly positive. – TonyK Apr 02 '20 at 15:42
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2It is not true for $n = 1$ – Damien Apr 02 '20 at 15:45
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If we consider the $x\geq 0$, we can easily conclude that $$f(x)=x^x-(x+1)^{x-1}=0$$ for $x=0 \vee x=1$. Now, if $x\geq1$, the function $f(x)$ is an increasing functiuon, so it's always positive. When we set $x=\frac{1}{2}$, we have: $$f\left(\frac{1}{2}\right)=\sqrt{\frac{1}{2}}-\sqrt{\frac{2}{3}}<0$$ Also, you can't $x$ to be negative because if $x$ is in the form $-\frac{1}{2k}, k\in N_0$, the expression $x^x$ is not defined.
Now, we consider $x=-3$, we have: $$f(-3)=-\frac{1}{27}-\frac{1}{16}=-\frac{43}{432}<0$$ So, if $x<0$ the function $f(x)$ isn't always positive.
Matteo
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