Prove that $\left(\frac{x^n+1}{x^{n-1}+1}\right)^n+\left(\frac{x+1}{2}\right)^n\geq x^n+1$ 
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!
 A: Mathematica confirms that for concrete values of $n$, e.g.
n = 30; ForAll[x,x > 0, ((x^n+1)/(x^(n-1)+1))^n + ((x + 1)/2)^n >= x^n + 1];Resolve[%,Reals]


True

Mathematica fails with it   in the general case of an arbitrary positive integer $n$.
A: Corrected proof, see GH from MO's comment and answer: A generalization of the inequality gives more flexibility for variations of parameters, which eventually yields a proof. One observation is $\frac{x+1}{2}=\frac{x^b+1}{x^{b-1}+1}$ for $b=1$. If one wants to achieve $(\frac{x^a+1}{x^{a-1}+1})^n+(\frac{x^b+1}{x^{b-1}+1})^n\geq x^n+1$ for all $x\ge0$, looking at derivatives at $x=1$ shows that a necessary condition is $a+b-1\ge n$. In fact, the following generalization holds true for all real $a,b\ge1$ and $x\ge0$:
\begin{equation}
\left(\frac{x^a+1}{x^{a-1}+1}\right)^{a+b-1}+\left(\frac{x^b+1}{x^{b-1}+1}\right)^{a+b-1}\geq x^{a+b-1}+1.
\end{equation}
Setting $n=a+b-1$, $a=(n+1)/2+c/2$, $x=y^2$ and $z=y^c$, this inequality is equivalent to
\begin{equation}
\left(\frac{y^{n+1}+z}{y^{n-1}+z}\right)^n+\left(\frac{y^{n+1}+1/z}{y^{n-1}+1/z}\right)^n\geq y^{2n}+1,
\end{equation}
where $y>0$ is arbitrary and $z>0$ is in an interval depending on $y$ and $n$. However, the inequality holds true for all $y>0$ and $z>0$. By the $z\leftrightarrow\frac{1}{z}$ and $y\leftrightarrow\frac{1}{y}$ symmetries, we assume in the following $0<z,y\le1$.
Fix $y$ and $n$. We show that the left hand side is monotonically increasing in $z$. Taking that for granted the assertion follows, for we have equality for $z=0$.
Taking the derivative with respect to $z$, the monotonicity is equivalent to
\begin{equation}
\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}.
\end{equation}
(In a previous version, there was a miscalculation, observed by GH from MO, which made the rest of the "proof" easier.)
Raising both sides to the $\frac{1}{(n-1)(n+1)}$-th power, and fixing now $y$ and $z$, the claim follows once we know that
\begin{equation}
(0,\infty)\to\mathbb R,\;\;t\mapsto\left(\frac{y^t+z}{zy^t+1}\right)^{\frac{1}{t}}
\end{equation}
is monotonically increasing in $t$. For this GH from MO's answer contains an elegant proof. Here is another one:
Take the derivative (with respect to $t$) of the logarithm of this function, then multiply by $t^2$ and set $w=y^t$. The result is
\begin{equation}
h(w,z):=\log w\cdot w\cdot(\frac{1}{w+z}-\frac{z}{zw+1})+\log(zw+1)-\log(w+z).
\end{equation}
We need to show that $h(w,z)\ge0$ for all $w,z\in(0,1)$. The derivative of $h(w,z)$ with respect to $w$ is
\begin{equation}
\frac{\partial h(w,z)}{\partial w}=\frac{(1-w^2)(1-z^2)z\log w}{((zw+1)(w+z))^2}<0.
\end{equation}
So $h(w,z)$, for fixed $z$, is decreasing in $w$. Thus $h(w,z)\ge h(1,z)=0$.
A: 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.
