About a Conjecture: $-\left(\frac{x^{n}+1}{x^{n-1}+1}\right)^{n}-\left(\frac{x+1}{2}\right)^{n}+\left(x^{\frac{x}{x+1}}+\sqrt{x}-1\right)^{n}+1\leq 0$ Hi it's a conjecture wich refine for $0< x\leq 1$ the inequality Refinement of a famous inequality :
Problem/Conjecture
Let $0<x\leq 1$ then for $n\geq 3$ a natural number it seems we have the inequality :
$$f(x)=-\left(\frac{x^{n}+1}{x^{n-1}+1}\right)^{n}-\left(\frac{x+1}{2}\right)^{n}+\left(x^{\frac{x}{x+1}}+\sqrt{x}-1\right)^{n}+1\leq 0$$
Some material :
Here New bound for Am-Gm of 2 variables I have proved using Gerber's theorem , the inequality :
$$\left(x^{\frac{x}{x+1}}+\sqrt{x}-1\right)^{n}\geq x^n$$
And here https://mathoverflow.net/questions/337457/prove-that-left-fracxn1xn-11-rightn-left-fracx12-rightn the inequality in the first link is shown greatly .
Idea for a proof (weaker in fact):
It seems we have under the constraint above  :
$$\left(\frac{x^{n}+1}{x^{n-1}+1}\right)-\left(\frac{\left(n^{\frac{1}{n}}-1\right)}{n^{\frac{1}{n}}}\right)^{\frac{1}{n}}\geq 0$$
And :
$$\left(\frac{x^{n}+1}{x^{n-1}+1}\right)-\frac{\left(x+1\right)}{2}\geq 0$$
And for $x\in(0,0.5]$:
$$x^{\frac{x}{x+1}}+\sqrt{x}-1-\left(\frac{\left(n^{\frac{1}{n}}-1\right)}{n^{\frac{1}{n}}}\right)^{\frac{1}{n}}\le 0$$
And finally for $x\in(0,0.5]$ :
$$\left(\frac{x^{n}+1}{x^{n-1}+1}\right)+\frac{\left(x+1\right)}{2}-\left(x^{\frac{x}{x+1}}+\sqrt{x}-1\right)-\left(\frac{\left(n^{\frac{1}{n}}-1\right)}{n^{\frac{1}{n}}}\right)^{\frac{1}{n}}\ge0$$
So applying the Karamata's inequality on $f(x)=x^n$ with  $x\in[0,0.5]$ gives a weaker result if i'm not wrong .
Edit :
Using the bounds above  we have a bound for the derivative for $n\geq 6$ :
$$j\left(x\right)=-n\left(\frac{x^{2n}+1}{x^{2\left(n-1\right)}+1}\right)^{\left(n-1\right)}\cdot\frac{2x^{2n+1}\left(x^{2n}+n\left(x^{2}-1\right)+1\right)}{\left(x^{2n}+x^{2}\right)^{2}}-2^{-n}\cdot2n\cdot x\left(x^{2}+1\right)^{\left(n-1\right)}+n\left(x^{2}\right)^{\left(n-1\right)}\cdot\left(\left(2x^{2}\right)x\cdot\frac{\left(x^{2}+2\ln\left(x\right)+1\right)}{\left(x^{2}+1\right)^{2}}+1\right)\leq \frac{d}{dx}f\left(x^{2}\right)$$
How to (dis)prove it ?

Not an answer just a useful remark :
We can split the problem because it seems we have :
Let $\exists x\in(0,1]$ and $\exists n\geq 3$ a natural number such that
$$\frac{\left(1+\frac{1}{2^{n}}+\frac{1}{n^{2}}-nx\right)\left(x^{n}+1\right)^{n}}{\left(x^{\left(n-1\right)}+1\right)^{n}}+nx-\frac{1}{n^{2}}-\frac{1}{2^{n}}-1\geq 0$$
And :
$$\frac{\left(x+1\right)^{n}}{2^{n}}+\frac{\left(nx-\frac{1}{n^{2}}-\frac{1}{2^{n}}\right)\left(x^{n}+1\right)^{n}}{\left(x^{\left(n-1\right)}+1\right)^{n}}-nx+\frac{1}{2^{n}}+\frac{1}{n^{2}}-\left(x^{\frac{x}
{x+1}}+\sqrt{x}-1\right)^{n}\ge0$$
I go a little bit further :
Define :
$$f\left(x\right)=\frac{\left(nx-\frac{1}{n^{2}}-\frac{1}{2^{n}}\right)\left(x^{n}+1\right)^{n}}{\left(x^{\left(n-1\right)}+1\right)^{n}}-\left(n-\frac{1}{2^{n}}\right)x+1/n^{2}+\frac{1}{2^{n}},g(x)=\frac{\left(x+1\right)^{n}}{2^{n}}-\frac{1+2x}{2^{n}},h\left(x\right)=\frac{x}{2^{n}}-\left(\sqrt{x}+x^{\frac{x}{x+1}}-1\right)^{n}$$
Let $a,b\in(0,1)$ $\exists x\in(a,b]$ and $\exists n\geq 3$ a natural number such that :
$$h(x)>0,f(x)>0,g(x)>0$$


We can also study the inequality for $x,y,z\in[0.5,1]$ and $n\geq 10$:
$$0\le y^{n}+z^{n}-1-\left(x-1\right)^{n}$$
This inequality seems true for :
Let :
$r(x)=\frac{x^{n}+1}{x^{n-1}+1}$
Denotes by $d=x_{min}\in[0.5,1]$ the value such that :
$$r'(d)=0$$
Then taking $z\in[d-1/2^{n},1]$ and $x\in[0.5,1]$ and $y=r(x)$ the inequality generalized seems true .
I haven't a proof yet but with all this stuff we can show the inequality .

