Let $x_{n}$ be the positive real root of $(x^{n-1}+2^n)^{n+1} = (x^{n} + 2^{n+1})^{n}$, how to prove that $x_{n} > x_{n + 1}$? Let $x_{n}$ be the positive real root of equation
$$(x^{n-1}+2^n)^{n+1} = (x^{n} + 2^{n+1})^{n}$$
How to prove that $x_{n} > x_{n + 1}$?
Actually, $x_{n} > 2$ and I get that $x_{1} = 5, x_{2} \approx 3.5973, x_{3} \approx 3.1033$

replace $x$ by $2x$, we have
\begin{equation}
1+\frac{x^n}{2} = \Big(1+\frac{x^{n-1}}{2}\Big)^{1+\frac{1}{n}} > 1 + \big(1+\frac{1}{n}\big)\frac{x^{n-1}}{2}
\end{equation}
hence $x> 1+1/n$. Let
\begin{equation}
f(y) = 1+\frac{y^{n+1}}{2} - \Big(1+\frac{y^n}{2}\Big)^{1+\frac{1}{n+1}}
\end{equation}
then
\begin{equation}
f'(y) = \frac{y^n}{2}\Big(n+1-\frac{n(n+2)\big(1+\frac{y^n}{2}\big)^{\frac{1}{n+1}}}{(n+1)y}\Big)
\end{equation}
$f'(y)>0$ if $y>1+1/n$. It is easy to check $f(x)>0$, hence $x>y_0$(root of $f$).
