TITLE: Prove that $\left(\frac{x^n+1}{x^{n-1}+1}\right)^n+\left(\frac{x+1}{2}\right)^n\geq x^n+1$
QUESTION [27 upvotes]: Let $x>0$ and $n$ be a natural number. Prove that:
  $$\left(\frac{x^n+1}{x^{n-1}+1}\right)^n+\left(\frac{x+1}{2}\right)^n\geq x^n+1.$$

This question is very similar to many contests problems, but I think it's much more harder than contest problem and it's just impossible to solve this problem during a competition.
In my fifth comment I wrote:
I took it here:   https://math.stackexchange.com/questions/3304808/refinement-of-a-famous-inequality. 
but I think there is no a chance that it will be solved there. 
I tried to use $$M_n^2\geq M_{n+1}M_{n-1},$$ where $x>0$ and $M_n=\sqrt[n]{\frac{x^n+1}{2}}$, but without success. 
I think a perspective way to solve this problem it's something like the following wrong way:
https://mathoverflow.net/revisions/337457/1
Thank you!

REPLY [12 votes]: This is a supplement (correction) to Peter Mueller's nice solution. As he observed, it suffices to show that, for any fixed $n\geq 1$ and $y\in[0,1]$, the function
$$z\mapsto\left(\frac{y^{n+1}+z}{y^{n-1}+z}\right)^n+\left(\frac{zy^{n+1}+1}{zy^{n-1}+1}\right)^n,\qquad z\in(0,1),$$
is increasing. (Indeed, $y:=x^{1/2}$ and $z:=x^{(n-1)/2}$ yields the LHS of the OP's inequality, while $y:=x^{1/2}$ and $z:=0$ yields the RHS of the OP's inequality.) Taking the derivative with respect to $z$, the statement becomes
$$\left(\frac{y^{n+1}+z}{zy^{n+1}+1}\right)^{n-1}\ge \left(\frac{y^{n-1}+z}{zy^{n-1}+1}\right)^{n+1},\qquad y,z\in(0,1).$$
Let us now fix $y,z\in(0,1)$ and think of $n\geq 1$ as the variable. Taking the logarithm of both sides and dividing by $(n-1)(n+1)$, it suffices to show that the function
$$t\mapsto\frac{1}{t}\log\frac{y^t+z}{zy^t+1},\qquad t>0,$$
is increasing. Making the change of variable $w:=y^t$, it suffices to show that the function
$$w\mapsto\frac{\log(w+z)-\log(wz+1)}{\log w},\qquad w\in(0,1),$$
is increasing. Writing $w=:\tanh u$ and $z=:\tanh v$, it suffices to show that the function
$$u\mapsto\frac{\log\tanh(u+v)}{\log\tanh(u)},\qquad u>0,$$
is increasing. Taking the derivative with respect to $u$, the statement becomes
$$\sinh(u)\cdot\cosh(u)\cdot\log\tanh(u)\geq\sinh(u+v)\cdot\cosh(u+v)\cdot\log\tanh(u+v).$$
That is, it suffices to show that the function
$$u\mapsto \sinh(u)\cdot\cosh(u)\cdot\log\tanh(u),\qquad u>0,$$
is decreasing. With the notation $s:=-\log\tanh(u)$, we have
$$\sinh(u)\cdot\cosh(u)\cdot\log\tanh(u)=\frac{e^{-s}}{\sqrt{1-e^{-2s}}}\cdot\frac{1}{\sqrt{1-e^{-2s}}}\cdot(-s)=\frac{-s}{2\sinh s},$$
hence it suffices to show that the function
$$s\mapsto\frac{\sinh s}{s},\qquad s>0$$
is increasing. However, this is clear, because the Taylor series of this function converges everywhere, and it has nonnegative coefficients.