Prove that $\left(\frac{x^n+1}{x^{n-1}+1}\right)^n+\left(\frac{x+1}{2}\right)^n\geq x^n+1$

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 $h(w,z)\ge h(1,z)=0$.

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.

$\newcommand{\s}{\overset{\text{sgn}}=} \newcommand{\Dx}{\text{Dx}} \newcommand{\logDx}{\text{logDx}} \newcommand{\DlogDx}{\text{DlogDx}} \newcommand{\DDDlogDx}{\text{DDDlogDx}} \newcommand{\DDDDDlogDx}{\text{DDDDDlogDx}} \newcommand{\dif}{\text{dif}} \newcommand{\Ddif}{\text{Ddif}} \newcommand{\R}{\mathbb{R}}$ Let us show that the inequality in question holds for all real $n\ge5$; the cases when $n\in\{1,2,3,4\}$ are verified directly. By a comment of Pietro Mayer, without loss of generality $0<x<1$. We shall reduce the problem to the completely algorithmic problem of checking sign patterns of several polynomials in $n,x$, of total degrees $\le11$. This reduction is done in a few steps:

Step 1: Eliminating $(\frac{1+x}2)^n$: The inequality in question can be rewritten as \begin{equation} u(x):=u_n(x):=n \ln \left(\frac{x^n+1}{x^{n-1}+1}\right) -\ln \left(x^n+1-z^n\right)\ge0, \end{equation} where $z:=z_x:=\frac{1+x}2$. Note that \begin{multline*} u'(x)\frac{x (1+x)}n \left(x^{1-n}+x^n+x+1\right) \left(x^n+1-z^n\right) \\ =\Dx:=\left(n \left(1-x^2\right)+\left(x^{2-n}-1\right) \left(1+x^n\right)\right) z^n-(n-1) \left(1-x^2\right) \left(1+x^n\right), \end{multline*} so that \begin{equation} u'(x)\s\Dx\s\logDx(x), \end{equation} where $\s$ denotes the equality in sign and
\begin{equation} \logDx(x):=\logDx_n(x):=n \ln z-\ln \frac{(n-1) \left(1-x^2\right) \left(1+x^n\right)}{n \left(1-x^2\right)+\left(x^{2-n}-1\right) \left(1+x^n\right)}. \end{equation} Here and in the sequel, $\Dx$, $\logDx$, etc. are atomic, "indivisible" symbols; $\Dx$ refers to the derivative (of $u$) in $x$, $\logDx$ refers to a certain kind of logarithmic modification of $\Dx$, etc. Next, let \begin{multline*} \DlogDx(x):=\DlogDx_n(x):= \\ \logDx'(x)(1-x) (1+x) x^{n-1} \left(1+x^n\right) \left(n \left(1-x^2\right)+\left(x^{2-n}-1\right) \left(1+x^n\right)\right) \\ =n^2 (x-1)^2 (x+1) \left(x-x^n\right) x^{n-2}-2 \left(x^n-1\right) \left(x^n+1\right)^2+\frac{n (x-1) \left(x^n+1\right)^2 \left(x^n+x\right)}{x}. \end{multline*} So, we get a polynomial in $x^n$ of degree $3$ over the field $\R(n,x)$ of all real rational functions in $n,x$.

Step 2: Reducing the degree from $3$ to $2$: Let \begin{multline*} \DDDlogDx(x):= \DlogDx''(x) x^{3 - 3 n}\\ =x^{3-3 n} (n (n x-n+2 x+2) (n^2 x^2-n^2+n x^2+2 n-1) x^{n-3} \\ -2 (n-1) n (x-1) (2 n^2 x^2-2 n^2+n x^2-2 n x+3 n+2 x) x^{2 n-4}+n (3 n-1) (3 n x-3 n-6 x+2) x^{3 n-3}) \\ \s\DlogDx''(x). \end{multline*} Taking the second derivative $\DlogDx''(x)$ of the polynomial $\DlogDx(x)$ in $x^n$ of over $\R(n,x)$ kills the free term of that polynomial. Thus, we get the polynomial $\DDDlogDx(x)$ of degree $2$ in $x^{-n}$ of over $\R(n,x)$.

Step 3: Reducing the degree from $2$ to $1$: Let \begin{equation} \DDDDDlogDx(x):= \frac{\DDDlogDx''(x)}{2 (n - 1) n^2 x^{-3 - 2 n}} = A_n(x) - x^n B_n(x), \end{equation} \begin{equation} A_n(x):=\left(2 n^3+3 n^2-5 n-6\right) x^4+\left(-2 n^3+3 n^2+3 n-2\right) x^3+\left(-2 n^3-n^2+5 n-2\right) x^2+\left(2 n^3-5 n^2+n+2\right) x, \end{equation} \begin{equation} B_n(x):=2 n^3+3 n^2-\left(-2 n^3+5 n^2-n-2\right) x^3-\left(2 n^3+n^2-5 n+2\right) x^2-\left(2 n^3-3 n^2-3 n+2\right) x-5 n-6, \end{equation} so that \begin{equation} \DDDlogDx''(x)\s A_n(x) - x^n B_n(x). \end{equation} Thus, we get the polynomial $\DDDDDlogDx(x)$ of degree $1$ in $x^n$ of over $\R(n,x)$.

Step 4: Reducing the degree from $1$ to $0$:
We can see that (under the conditions $n\ge5$ and $0<x<1$, assumed everywhere here) $B_n(x)>0$. So, $\DDDlogDx''(x)<0$ whenever $A_n(x)\le0$.

Further, let \begin{equation} \dif(x) = \dif_n(x) :=\ln\frac{A_n(x)}{B_n(x)} - n \ln x\s A_n(x) - x^n B_n(x)\s \DDDlogDx''(x) \end{equation} wherever $A_n(x)>0$, and then \begin{multline*} \Ddif(x) = \Ddif_n(x) :=\dif'(x)\frac{A_n(x)B_n(x)}{(n+1)(n-2)} \\ =-4 n^5 (x-1)^4 (x+1)^2+4 n^4 (x-1)^4 (x+1)^2+n^3 (x-1)^2 \left(15 x^4+16 x^3-10 x^2+16 x+15\right) \\ -4 n^2 \left(x^2-1\right)^2 \left(5 x^2+x+5\right)+n \left(-x^6+30 x^5+41 x^4-44 x^3+41 x^2+30 x-1\right) \\ +2 \left(3 x^6-6 x^5-11 x^4-36 x^3-11 x^2-6 x+3\right) \\ \s\dif'(x), \end{multline*} finally getting a polynomial in $n,x$.

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Now we need to trace the above steps back:

Looking back at the polynomial $A_n(x)$, (for $x\in(0,1)$) we find that $A_n(x)\le0$ iff $x_1\le x\le x_2$, where $x_1=x_1(n)$ and $x_2=x_2(n)$ are the two roots of $A_n(x)$ in $(0,1)$ such that $x_1<x_2$.

Further, $\Ddif<0$ and hence $\dif'<0$ on $(0,x_1]$; and $\Ddif>0$ and hence $\dif'>0$ on $[x_2,1)$. So, $\dif$ decreases on $(0, x_1]$ and increases on $[x_2, 1)$. So, $\dif$ is $+-$ on $(0, x_1]$ (that is, $\dif$ can switch sign at most once on $(0, x_1]$, and only from $+$ to $-$). Similarly, $\dif$ is $-+$ on $[x_2, 1)$.

But also $\dif(1)=0$. So, actually $\dif<0$ on $[x_2, 1)$.

So, $\DDDlogDx''$ is $+-$ on $(0, x_1]$ and $\DDDlogDx'' < 0$ on $[x_2, 1)$.

Also, $A < 0$ and hence $\DDDlogDx'' < 0$ on $[x_1, x_2]$. So, $\DDDlogDx''$ is $+-$ on $(0, 1)$. So, $\DDDlogDx$ is convex-concave on $(0, 1)$. Also, $\DDDlogDx(1)=0$.

So, $\DDDlogDx$ is $+-+$ on $(0, 1)$. So, $\DlogDx$ is convex-concave-convex on $(0, 1)$. Also, $\DlogDx(1)=\DlogDx'(1)=\DlogDx''(1)=0>-8n(n^2-1)=\DlogDx'''(1)$ and $\DlogDx(0+)=2-n<0$. So, $\DlogDx$ is $-+$; so, $\logDx$ is decreasing-increasing.

Also, $\logDx(1-)=0$. So, $\logDx$ is $+-$, and hence so is $\Dx$ (with $z = \frac{1 + x}2$).

Recalling that $u'(x)\s\Dx$, we see that $u_n(x)$ is increasing-decreasing (in $x\in(0,1)$). Also, $u_n(0)=-\ln(1 - 2^{-n})>0$ and $u_n(1)=0$.

Thus, $u>0$, which concludes the proof.