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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, $n\geq1$. Prove that: $$\left(\frac{x^n+1}{x^{n-1}+1}\right)^n+\left(\frac{x+1}{2}\right)^n\geq x^n+1.$$

My attempts:

For all reals $x>0$ and $n>1$ we have: $$\frac{x^n+1}{x^{n-1}+1}+\frac{x+1}{2}\geq x+1$$ it's $$2(x^n+1)\geq(x+1)(x^{n-1}+1),$$ which is true by Chebyshov because $(x,1)$ and $(x^{n-1},1)$ have the same ordering.

Also, $$\frac{x^n+1}{x^{n-1}+1}\cdot\frac{x+1}{2}\geq x\cdot1$$ it's $$x^{n+1}+1\geq x^n+x,$$ which is true by Murhead because $(n+1,0)\succ(n,1).$

Also, in the inequality $$\frac{x^n+1}{x^{n-1}+1}\geq\frac{x+1}{2}$$ the equality occurs for $x=1$ only, for which the starting inequality is obviously true.

Thus, it's enough to prove our inequality for $\frac{x^n+1}{x^{n-1}+1}\neq\frac{x+1}{2}.$

Now, let $\frac{x^n+1}{x^{n-1}+1}=a$, $\frac{x+1}{2}=b$, $x=c$ and $1=d$.

Thus, $$a+b\geq c+d,$$ $$ab\geq cd$$ and we need to prove that $$a^n+b^n\geq c^n+d^n,$$ where $a\neq b$ and $c\neq d$.

Indeed, let $a+b=u$ and $ab=v$.

Thus, since $a^n+b^n$ is a symmetric expression of $a$ and $b$, we can think that $$a^n+b^n=f(u,v)$$ and it's enough to prove that $$\frac{\partial f}{\partial u}\geq0$$ and $$\frac{\partial f}{\partial v}\geq0.$$ From $a+b=u$ we obtain: $$\frac{\partial a}{\partial u}+\frac{\partial b}{\partial u}=1$$ and from $ab=v$ we obtain: $$\frac{\partial a}{\partial u}b+\frac{\partial b}{\partial u}a=0,$$ which gives $$\frac{\partial a}{\partial u}=\frac{a}{a-b}$$ and $$\frac{\partial b}{\partial u}=\frac{b}{b-a}.$$ Id est, $$\frac{\partial f}{\partial u}=\frac{\partial f}{\partial a}\frac{\partial a}{\partial u}+\frac{\partial f}{\partial b}\frac{\partial b}{\partial u}=\frac{na^n}{a-b}+\frac{nb^n}{b-a}=\frac{n(a^n-b^n)}{a-b}>0.$$ Now, $$\frac{\partial a}{\partial v}+\frac{\partial b}{\partial v}=0$$ and $$\frac{\partial a}{\partial v}b+\frac{\partial b}{\partial v}a=1,$$ which gives $$\frac{\partial a}{\partial v}=\frac{-1}{a-b}$$ and $$\frac{\partial b}{\partial v}=\frac{1}{a-b}.$$ But $$\frac{\partial f}{\partial v}=\frac{\partial f}{\partial a}\frac{\partial a}{\partial v}+\frac{\partial f}{\partial b}\frac{\partial b}{\partial v}=-\frac{n(a^{n-1}-b^{n-1})}{a-b}<0$$ and we did not get a proof.

Help me please to prove the starting inequality. Thank you!