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Problem :

Let $50\geq a>\varepsilon>0 $,$x>\varepsilon>0$ then it seems we have :

$$g(x)=2\left(\frac{\left(ax\right)^{\frac{1}{ax-1}}}{e\ln\left(\left(ax\right)^{\frac{1}{ax-1}}\right)}\right)^{\frac{a}{2\left(a+1\right)}}\leq \sqrt{\frac{x^{a}}{a}}+\sqrt{\frac{a^{x}}{x}}=f(x)$$



Using Generalized Young inequality where $b=1$ we have :

$$h(x)=\left(\frac{1}{a^{b}}+\frac{1}{x^{b}}\right)\left(a^{\sqrt{2^{-1}}x^{\left(1-b\right)}}\cdot x^{\left(\sqrt{2^{-1}}\left(a\right)^{\left(1-b\right)}\right)}\cdot a^{\frac{\left(b-0.5\right)\sqrt{2}}{a^{b}}}\cdot x^{\frac{\left(b-0.5\right)\sqrt{2}}{x^{b}}}\right)^{\frac{1}{\sqrt{2}a^{-b}+\frac{\sqrt{2}}{x^{b}}}}\leq f(x)$$

It seems for $a\leq 1$ that :

$$h(x)\geq g(x)$$

It's not true for $a\geq 1$

After taking the logarithm on both side it seems we have a refinement of : Prove $2(x + y)\ln \frac{x + y}{2} - (x + 1)\ln x - (y + 1)\ln y \ge 0$



Edit : We have the inequality on $(0,1]$ :

$$\frac{\left(x+1\right)}{\left(x-1\right)}\ln\left(1+x-\sqrt{x}\right)\geq x\frac{\ln\left(x\right)}{x-1}$$

A proof can be found here New bound for Am-Gm of 2 variables

Edit 2 :

We have $x>0$:

$$\frac{\left(x-1\right)}{2}-\frac{\left(x-1\right)^{3}}{48}\leq \sqrt{x}\ln\sqrt{x}\leq \frac{\left(x-1\right)}{2}+\frac{\left(x^{-1}-1\right)^{3}}{48}$$

Edit 3 :

We have the easy inequality for $1\leq x\leq 3$ :

$$\frac{x^{\frac{1}{x-1}}}{e\ln x^{\frac{1}{x-1}}}\leq \frac{x^{-0.5}}{\ln\left(x^{-0.5}e\right)}$$

Last edit :

Let $x>1$ then we have :

$$\ln\left(\ln\left(x\right)\right)\leq \frac{x-e+\frac{\left(x-e\right)^{3}}{6e^{2}}}{x}$$

For $x>\frac{5}{100}$ we have the inequality :

$$m(x)=2\left(\frac{3}{4}\cdot\frac{1}{4}\left(\frac{2\left(x-1\right)}{x+1}+\frac{\left(x-1\right)}{\sqrt{x}}\right)+\left(\sqrt{\sqrt{\sqrt{x}}}-1\right)\right)\geq \ln x$$

And for $0<x\leq 1 $ :

$$p(x)=\frac{1}{2}\left(\frac{2}{3+x}\left(\frac{2\left(x-1\right)}{x+1}+\frac{\left(x-1\right)}{\sqrt{x}}\right)+4\left(\sqrt{\sqrt{x}}-1\right)\right)\leq \ln x$$

Let $\frac{5}{100}<x\leq 1$

$$\left(\frac{16}{10}\cdot p\left(x^{\frac{1}{8}}\right)+\frac{64}{10}\cdot m\left(x^{\frac{1}{8}}\right)\right)\geq \ln(x)$$

All the inequalities can be proved using derivatives .

The goal is to come to an inequality like :

$$k(a+x,ax,0)\geq 0$$

And then apply $uvw's$ method.



Question :

How to (dis)prove the problem ?

Reference :

Refined Young inequalities with Specht's ratio S Furuichi - Journal of the Egyptian Mathematical Society, 2012 - Elsevier

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Partial answer using conjectures:

Let $a,x\geq1$ then define :

$$\sqrt{\frac{x^{a}}{a}}+\sqrt{\frac{a^{x}}{x}}=f(x)$$

Now substitute $x=e^y$, then we have :

$$(f(e^y))''>0$$

See Wolfram alpha for a direct proof of this fact :

https://www.wolframalpha.com/input/?i=second+derivative+sqrt%28e%5E%28ax%29%2Fa%29%2Bsqrt%28a%5E%28e%5Ex%29%2Fe%5Ex%29

So by Jensen's inequality we have :

$$f(x)\geq 2f(\sqrt{ax})-f(a)$$

The OP's inequality have the form :

$$g(a,ax)\geq 0$$

Now denotes by :

$$n(x)=-2\sqrt{\frac{x^{x}}{x}}+2\left(\sqrt{\frac{x^{\sqrt{b}}}{\sqrt{b}}}+\sqrt{\frac{\left(\sqrt{b}\right)^{x}}{x}}\right)-2\left(\frac{b^{\frac{1}{b-1}}}{e\ln\left(b^{\frac{1}{b-1}}\right)}\right)^{\frac{x}{2\left(x+1\right)}}$$

Where $b=ax$

Conjectures :


Then for $1\leq b\leq 5$,$b\neq1$ and $x>0$ it seems we have :

$$\frac{d}{dx}\left(\frac{d}{dx}n\left(x\right)\right)\leq 0$$

Using this pre-fact the chord starting at the point $(1,n(1))$ and ending at $(\sqrt{ax},n(\sqrt{ax}))$ seems to have a positive slope so we just need to have :

$$n(1)\geq 0$$

A similar result can be done for $a,x\leq 1$


Edit 21/11/2021:

It seems we have better result using :

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

Following that we have two interesting function :

$$u(x)=2\sqrt{\frac{\left(\frac{1}{\sqrt{ax}}\right)^{a}}{a}}-\sqrt{\frac{\left(\frac{1}{a}\right)^{a}}{a}}-1+\ln\left(\left(ax\right)^{\frac{1}{2}}\right)$$

And

$$v(x)=2\sqrt{\sqrt{ax}\cdot a^{\frac{1}{\sqrt{ax}}}}-\sqrt{a^{\left(1+\frac{1}{a}\right)}}-1-\ln\left(\left(ax\right)^{\frac{1}{2}}\right)$$

Last remark 22/11/2021:

Now there is a possibility to use the Jensen's inequality for strong convex function (see the comment above)

Last edit 22/11/2021 :

We need to use strong convexity so we have :

$$h(x)=\frac{d}{dx}\left(\frac{d}{dx}f\left(x\right)\right)=\frac{a^{\frac{x}{2}}\left(x\ln\left(a\right)\left(x\ln\left(a\right)-2\right)+3\right)+\left(a^{\frac{3}{2}}-2\sqrt{a}\right)x^{\frac{\left(a+1\right)}{2}}}{4x^{\frac{5}{2}}}$$

Let $a\in[0.6,1]$ and $1\leq x\leq \frac{7}{3}$:

Then it seems we have :

$$h(x)\geq \frac{a^{\frac{1}{2}}3+\left(a^{3}-2a\right)x}{4ax^{\frac{5}{2}}}$$



We can go further reattaching the result above with Am-Gm using the facts that 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 :

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

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  • 1
    $\begingroup$ Why make it a community post? $\endgroup$ Commented Nov 20, 2021 at 17:52
  • $\begingroup$ @TymaGaidash because there is a serial down voter so... $\endgroup$ Commented Nov 22, 2021 at 13:55
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    $\begingroup$ (+1) @Erik Satie You could always try to flag/ make a meta post about it, or not. I like how you try to solve your own questions. $\endgroup$ Commented Nov 22, 2021 at 21:35

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