About a Conjecture: $-\left(\frac{x^{n}+1}{x^{n-1}+1}\right)^{n}-\left(\frac{x+1}{2}\right)^{n}+\left(x^{\frac{x}{x+1}}+\sqrt{x}-1\right)^{n}+1\leq 0$

Question

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 ?

Answer 1

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 .

---

Answer 2

This is not an answer - Just a few elements

$$f(x)=-\left(\frac{x^{n}+1}{x^{n-1}+1}\right)^{n}-\left(\frac{x+1}{2}\right)^{n}+\left(x^{\frac{x}{x+1}}+\sqrt{x}-1\right)^{n}+1$$ gives $f(0)=-2^{-n}$ and $f(1)=0$. We also have $f'(0)=-n\, 2^{-n} <0 $ and $f'(1)=0^-$

When $n$ is large, we have $$f(x)=-\frac{n \left(n^3-n-1\right) }{192 }(1-x)^4+O\left((x-1)^5\right)$$ which is always negative.

So, $f(x)$ starts decreasing and goes through a minimum value. We can approximate where the minimum occurs using the series expansion of $f'(x)$ and series reversion to get $$x_*=1-\frac{8 \left(n^3-n-1\right)}{5 \left(n^4-4 n^3-n^2+2 n+5\right)}-$$ $$\frac{64 \left(n^3-n-1\right)^2 \left(n^5-60 n^4+160 n^3+30 n^2+19 n-246\right)}{625 \left(n^4-4 n^3-n^2+2 n+5\right)^3}+\cdots$$

The problem is that we need a lot of terms before arriving at $$x_* \sim 1-\frac{4}{n}+\frac{7}{2n^2}+O\left(\frac{1}{n^3}\right)$$ which does not seem to be too bad $$\left( \begin{array}{ccc} n & \text{estimate} & \text{solution}\\ 3 & 0.055556 & 0.079320 \\ 4 & 0.218750 & 0.228742 \\ 5 & 0.340000 & 0.346682 \\ 6 & 0.430556 & 0.435779 \\ 7 & 0.500000 & 0.504355 \\ 8 & 0.554688 & 0.558444 \\ 9 & 0.598765 & 0.602076 \\ 10 & 0.635000 & 0.637962 \\ 20 & 0.808750 & 0.810192 \\ 30 & 0.870556 & 0.871505 \\ 40 & 0.902188 & 0.902894 \\ 50 & 0.921400 & 0.921962 \\ 60 & 0.934306 & 0.934773 \\ 70 & 0.943571 & 0.943971 \\ 80 & 0.950547 & 0.950895 \\ 90 & 0.955988 & 0.956297 \\ 100 & 0.960350 & 0.960628 \\ 200 & 0.980088 & 0.980226 \\ 300 & 0.986706 & 0.986797 \\ 400 & 0.990022 & 0.990091 \\ 500 & 0.992014 & 0.992069 \\ 600 & 0.993343 & 0.993389 \\ 700 & 0.994293 & 0.994332 \\ 800 & 0.995005 & 0.995040 \\ 900 & 0.995560 & 0.995590 \\ 1000 & 0.996004 & 0.996031 \\ 2000 & 0.998001 & 0.998015 \\ 3000 & 0.998667 & 0.998676 \\ 4000 & 0.999000 & 0.999007 \\ 5000 & 0.999200 & 0.999206 \\ 6000 & 0.999333 & 0.999338 \\ 7000 & 0.999429 & 0.999433 \\ 8000 & 0.999500 & 0.999503 \\ 9000 & 0.999556 & 0.999559 \\ 10000 & 0.999600 & 0.999603 \end{array} \right)$$

Edit

We also have $$f''(0)=-n(n-1)\,2^{-n} \qquad \text{and} \qquad f''(1)=0^-$$ As soon as $n>3$, there are two inflection points.