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Aug 10, 2019 at 20:00 history edited Peter Mueller CC BY-SA 4.0
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Aug 10, 2019 at 18:28 vote accept Michael Rozenberg
Aug 10, 2019 at 18:28 comment added Michael Rozenberg It's an extremely nice proof! Thank you Peter! @GH from MO Thank you!
Aug 10, 2019 at 11:49 comment added GH from MO @MichaelRozenberg: Peter Mueller proved the inequality contained in his first display. Setting $a:=n$ and $b:=1$ in this inequality, you obtain the inequality in your post. Alternatively, start from his second display, and set $y:=x^{1/2}$ and $z:=x^{(n-1)/2}$ there.
Aug 10, 2019 at 11:46 comment added GH from MO @PeterMueller: Nice!
Aug 10, 2019 at 11:18 history edited Peter Mueller CC BY-SA 4.0
added 2 characters in body
Aug 10, 2019 at 11:15 comment added Peter Mueller @GHfromMO Thank you for spotting the miscalculation and fixing the proof! I added another argument for the final monotonicity assertion.
Aug 10, 2019 at 11:13 history edited Peter Mueller CC BY-SA 4.0
Fixed the proof
Aug 10, 2019 at 10:17 comment added Michael Rozenberg $\frac{x+1}{2}=\frac{x^b+1}{x^{b-1}+1}$ it's $\left(x^{b-1}-1\right)(x-1)=0,$ which for $b\neq1$ gives $x=1$ only.
Aug 10, 2019 at 4:18 comment added GH from MO On the right hand side of your third display, a factor of $\frac{1}{z^2}$ is missing. That is, monotonicity is equivalent to $\left(\frac{y^{n+1}+z}{zy^{n+1}+1}\right)^{n-1}\ge \left(\frac{y^{n-1}+z}{zy^{n-1}+1}\right)^{n+1}$. I added below a detailed proof of this inequality, for any $n\in[1,\infty)$ and $y,z\in[0,1]$.
Aug 9, 2019 at 23:19 history edited Peter Mueller CC BY-SA 4.0
fixed a typo
Aug 9, 2019 at 22:00 history edited Peter Mueller CC BY-SA 4.0
completed the proof
Aug 9, 2019 at 21:55 history edited Peter Mueller CC BY-SA 4.0
completed the proof
Aug 9, 2019 at 15:27 history edited Peter Mueller CC BY-SA 4.0
added 119 characters in body
Aug 9, 2019 at 14:39 history edited Peter Mueller CC BY-SA 4.0
fixed a wrong statement about monotonicity
Aug 6, 2019 at 13:39 history answered Peter Mueller CC BY-SA 4.0