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 $x_{1} = 5, x_{2} \approx 3.5973, x_{3} \approx 3.1033$
 A: 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 $f$).
A: This is not a proof.
Consider that you look for the non-trivial positive zero of function
$$f_n(x)=(x^{n-1}+2^n)^{n+1} - (x^{n} + 2^{n+1})^{n}$$ If you compute the series of $\frac{f_n(x)}{x^n}$ to $O(x)$, you have
$$\frac{f_n(x)}{x^n}=\frac {a_n}x-b_n+O(x)\implies x_{(n)}=\frac{b_n}{a_n}$$ This gives as an approximation
$$x_{(n)}=2+\frac 2n$$ which is not fantastic but "almost" decent (I hope).
$$\left(
\begin{array}{ccc}
  n & \text{approximation} & \text{exact} \\
 10 & 2.20000 &  2.35954 \\
 20 & 2.10000 &  2.18495 \\
 30 & 2.06667 &  2.12467 \\
 40 & 2.05000 &  2.09405 \\
 50 & 2.04000 &  2.07552 \\
 60 & 2.03333 &  2.06309 \\
 70 & 2.02857 &  2.05417 \\
 80 & 2.02500 &  2.04747 \\
 90 & 2.02222 &  2.04224 \\
 100 & 2.02000 &  2.03805
\end{array}
\right)$$
