About an inequality wich is an upper bound for Am-Gm.

Question

Hi it's a upper bound for New bound for Am-Gm of 2 variables :

Problem :

Let $0<a\leq 1$ and $x\geq 1$ such that $2\leq a+x\leq 5$ then (dis)prove :

$$\left(1+\frac{1}{x+a}\left(\frac{4\left(x+a\right)^{2}}{\left(x+a+2\right)^{2}}-1\right)\right)\left(x+a-\frac{2\left(x+a\right)}{\left(x+a+2\right)}\right)-\left(ax\right)^{\frac{1}{x+a}}x^{\frac{x}{x+a}}a^{\frac{a}{a+x}}\geq0$$

---

---

In the link above you can find a proof for the lower bound .

If true i have tried to use Bernoulli's inequality wich is more than unsufficient .

Edit : Playing with Taylor series and improving it manually we have :

$$x+\ln\left(ax\right)-a\ln\left(x\right)+a\ln\left(a\right)+\frac{1-a}{x}+\frac{1}{92.95}\left(\frac{1-a}{x}\right)^{2}\simeq \left(ax\right)^{\frac{1}{a+x}}x^{\frac{x}{a+x}}a^{\frac{a}{a+x}}$$

For $0.75\leq a \leq 1$ and $x\geq 1$ wich is a good approximation but in reality unsufficient .

Edit 2:

Let $1\leq x\leq 2$ and $0.5\leq a \leq 1$ then it seems we have :

$$\frac{d}{dx}\left(\frac{d}{dx}g\left(x\right)\right)\geq 0$$

Where :

$$g\left(x\right)=\left(ax\right)^{\frac{1}{x+a}}x^{\frac{x}{x+a}}a^{\frac{a}{a+x}}$$

We can try to show the convexity using :

for $a,x>0$:

$$2\left(a+x\right)\ln\frac{a+x}{2}-\left(a+1\right)\ln a-\left(x+1\right)\ln x\geq 0$$

And :

$$a\ln a+x\ln x-\left(a+x\right)\ln\left(a+x-\sqrt{ax}\right)\geq 0$$

---

---

Using this supposition and using a chord on $g(x)$ we have for $x\in[1,2-a]$ and $0.5\leq a\leq 1$ :

$$g(x)\leq x+a-1$$

The rest seems easy .

---

---

Last edit :

Using directly differentiation and logarithm and multiplying by $x+a$:

We have using WA :

$$\frac{\left(12-7\left(a+x\right)^{2}\right)}{\left(a+x\right)^{2}\left(a+x+7\right)-4}+\frac{6}{a+x+2}+\ln\left(\frac{\left(a+x\right)\left(\left(a+x\right)^{2}\left(a+x+7\right)-4\right)}{\left(a+x+2\right)^{3}}\right)-\frac{1}{x}-\ln\left(x\right)$$

We can replace $-\frac{1}{x}$ by $\frac{-2}{a+x}$ and $-\ln\left(x\right)$ by $-\ln\left(0.5(x+a)\right)$ around the zero of the derivative

Now it's a single variable inequality in $x+a$

---

---

Question :

How to (dis)prove it ?

Answer 1

Partial answer :

We prove a first step :

we have for $x\in[1,2-a]$ and $0.5\leq a\leq 1$ :

$$g(x)\leq x+a-1$$

or we need to show :

$$f(x)=\left(a+1\right)\ln\left(a\right)+\left(x+1\right)\ln\left(x\right)-\left(a+x\right)\ln\left(a+x-1\right)$$

It's easy using a computer or WA to see that the function is convex on the interval above because :

$$f''(x)=\frac{(a-1)(a\left(x-1\right)+x^{2}-3x+1)}{something positive}\geq 0$$

So using a chord we have :

$$f(1)=0\quad, f(2-a)\leq 0$$

Again it's easy to show it

So using the convexity we have :

$$f(x)\leq \frac{\left(f\left(1\right)-f\left(2-a\right)\right)}{1-\left(2-a\right)}\left(x-1\right)+f\left(1\right)\leq 0$$

Now we need to show for $x\in[1,2-a]$ and $0.5\leq a\leq 1$:

$$\left(1+\frac{1}{x+a}\left(\frac{4\left(x+a\right)^{2}}{\left(x+a+2\right)^{2}}-1\right)\right)\left(x+a-\frac{2\left(x+a\right)}{\left(x+a+2\right)}\right)-(x+a-1)\geq 0$$

Wich is a single variable inequality or a polynomial with an obvious root .

We are done in this case .

Edit :

Following the same way we can improve the proof for $x\in[1,2-a+\frac{1}{2}\left(a-1\right)^{2}]$ and $0.5\leq a\leq 1$

---

---

---

Curious fact :

Let $a=0.5$ and $x=e$ then define :

$$k(x)=\left(1+\frac{1}{x+a}\left(\frac{4\left(x+a\right)^{2}}{\left(x+a+2\right)^{2}}-1\right)\right)\left(x+a-\frac{2\left(x+a\right)}{\left(x+a+2\right)}\right)-\left(ax\right)^{\frac{1}{x+a}}x^{\frac{x}{x+a}}a^{\frac{a}{a+x}}$$

And :

$$j(x)=\frac{d}{dx}\left(\frac{d}{dx}\left(g\left(x\right)\right)\right)$$

Then :

$$j(e)-k(e)>0$$

---

---

---

---

It seems that the function :

$$k(x)+g(x)$$

Is convex for $x\in(1,3)$

We can repeat the same reasoning as above on differents interval include in $(1,3)$ until getting the minimum .

edit 2:

We have :

$$r(x+a)=\left(1+\frac{1}{x+a}\left(\frac{4\left(x+a\right)^{2}}{\left(x+a+2\right)^{2}}-1\right)\right)\left(x+a-\frac{2\left(x+a\right)}{\left(x+a+2\right)}\right)=\frac{(x+a)((x+a)^3+7(x+a)^2-4)}{(x+a+2)^3}$$

The second derivative is :

$$r''(x)=\frac{-36x^2+144x+48}{somethingpositive}$$

So solving the quadratics we see that the function $r(x)$ is convex on $x\in[1,4]$

Using the fact above (convexity) we can do the same thing at first taking :

$$k(x)=r'\left(2+b\right)\left(x+a-\left(2+b\right)\right)+r\left(2+b\right)$$

And $b=\left(2-a\right)\left(2.25-a\right)\left(a-1\right)^{2}$ where $0.5\leq a \leq 0.9$

on the interval $[2-a+1.5\left(a-1\right)^{2},c]$ where $c$ is :

$$k(c)-\left(ac\right)^{\frac{1}{c+a}}c^{\frac{c}{c+a}}a^{\frac{a}{a+c}}=0$$

We can choose the interval $x\in[2-a+0.5(a-1)^2,c]$ for $0.95\leq a\leq 0.999$ .With my previous work it remains to find the interval for $0.9\leq a\leq 0.95$

Now it includes the minimum on $x\in|1,2.5)$

...To be continued