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 $x_{1} = 5, x_{2} \approx 3.5973, x_{3} \approx 3.1033$
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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 $f$).
yisishoujo
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I have not got how to check $f(x)> 0$ – Blanco Sep 06 '20 at 17:50
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let $1+x^{n-1}/2 = \alpha^n$, then $1+x^{n}/2 = \alpha^{n+1}$, $x=(\alpha^{n+1}-1)/(\alpha^n-1)$, $f(x)$ is something times $\alpha +1/\alpha -2$ – yisishoujo Sep 07 '20 at 00:44
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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).
$$\left( \begin{array}{ccc} n & \text{approximation} & \text{exact} \\ 10 & 2.20000 & 2.35954 \\ 20 & 2.10000 & 2.18495 \\ 30 & 2.06667 & 2.12467 \\ 40 & 2.05000 & 2.09405 \\ 50 & 2.04000 & 2.07552 \\ 60 & 2.03333 & 2.06309 \\ 70 & 2.02857 & 2.05417 \\ 80 & 2.02500 & 2.04747 \\ 90 & 2.02222 & 2.04224 \\ 100 & 2.02000 & 2.03805 \end{array} \right)$$
Claude Leibovici
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