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This is a question related to this For $abc=1$ prove that $\sum\limits_{cyc}\frac{a}{a^{11}+1}\leq\frac{3}{2}.$ based on a new approach.

First I remark that it's equivalent to the inequality : $$\frac{x^{10}}{x^{11}+1}+\frac{y^{10}}{y^{11}+1}+\frac{z^{10}}{z^{11}+1} \leqslant \frac{3}{2}.$$

The antidervative of the function $f(x)=\dfrac{x^{10}}{x^{11}+1}$ is very very simple, i.e. $$F(x)=\frac{\ln(x^{11}+1)}{11}.$$

Based on this remark I have create two inequalities like this:

Let $x,y$ such as $x\ge 1$ and $y\ge 1$ and $x\ge y$ then for $x-y\le 0.1$ we have: $$\frac{x^{10}}{x^{11}+1}+\frac{y^{10}}{y^{11}+1}\leq \frac{(2+x-y)\ln\left(\frac{x^{11}+1}{y^{11}+1}\right)}{11(x-y)},$$

and

$$\frac{y^{10}}{y^{11}+1}+\frac{(\frac{1}{xy})^{10}}{(\frac{1}{xy})^{11}+1}\leq \frac{(1+x-y)\ln\left(\frac{y^{11}+1}{(\frac{1}{xy})^{11}+1}\right)}{22(x-y)}+x-y.$$

and $$\frac{x^{10}}{x^{11}+1}+\frac{(\frac{1}{xy})^{10}}{(\frac{1}{xy})^{11}+1}\leq \frac{\ln\left(\frac{x^{11}+1}{(\frac{1}{xy})^{11}+1}\right)}{22(x-y)}+y-x.$$

So do you have nice ideas to prove these two inequalities or counter-examples?

Thanks a lot.

Edit for MartinR :

First I think that even if I check the two inequalities it doesn't cancel the problem related to the veracity of them (I'm human) .Secondly , if I take 0.1 it just a numerical test because it corresponds to the inequality : $$ 0.5[\frac{(2+x-y)\ln\left(\frac{x^{11}+1}{y^{11}+1}\right)}{11(x-y)}+\frac{\ln\left(\frac{x^{11}+1}{(\frac{1}{xy})^{11}+1}\right)}{22(x-y)}+\frac{(1+x-y)\ln\left(\frac{y^{11}+1}{(\frac{1}{xy})^{11}+1}\right)}{22(x-y)}]\leqslant 1.5$$

Where I just sum the inequalities above and add the inequality with $y$ and $\frac{1}{xy}$.

Thirdly if all what I said is right it proved the most difficult part of the original inequality where the value of $x,y,z$ are closed to 1 .

Now the second axis of my proof is related to the inequality :

$$g(x)=f(x)+f(\frac{1}{x})+f(1)=\frac{x^{10}}{x^{11}+1}+\frac{x}{x^{11}+1}+0.5\leqslant \frac{3}{2}.$$

To prove it correctly we have the following theorem (that we can prove by contradiction)

Let $a,b,c,d$ be positive real numbers such as : $a\geq b$ and $c\geq d$ with : $a\geq c$ and $ab\geq cd$ then we have : $$a+b\geq c+d$$

For the case $0<x\leq 1$ we have :

$c=x\leq a=1$ and : $dc=x^{10}(x)\leq ab=x^{11}(1)$ So we have proved the inequality : $$x^{10}+x\leq x^{11}+1$$ Or : $$\frac{x^{10}+x}{x^{11}+1}\leq 1$$ Or :

$$\frac{x^{10}}{x^{11}+1}+\frac{x}{x^{11}+1}+0.5\leq 1.5$$

For the case $x\geq1 $ remark this :

$g(x)=g(\frac{1}{x})$

So the original inequality is proved for $(x)(\frac{1}{x})1=1$

Now the third axis is to prove the following inequality for $z\leq 1$ and $0.5\leq x\leq 1 $ and $y\geq 2$ with $$\frac{x^{10}}{x^{11}+1}+\frac{x}{x^{11}+1}+0.5\geq \frac{x^{10}}{x^{11}+1}+\frac{y^{10}}{y^{11}+1}+\frac{z^{10}}{z^{11}+1}$$

It's easy with the following inequalities : $0.5\geq \frac{z^{10}}{z^{11}+1}$ And $$\frac{y^{10}}{y^{11}+1}\leq f(2)\leq \frac{x}{x^{11}+1}$$

So we can continue like this but the case where $x,y,z$ are closed to 1 is very hard so MartinR are you agree with these inequalities and my reasoning ?

  • 2
    Would a proof of your inequalities help to solve the original problem in https://math.stackexchange.com/q/2602035/42969 (and how?), or why do you call it a “new strategy”? What is the relation between these problems? You ask for a “proof or counter-example” – but what do you think? Are there any reasons why those estimates should be true (or not)? Did you check them for specific values? How did you come up with these expressions? Why $0.1$? – Martin R Mar 01 '18 at 18:22
  • See my edits if you have other question I'm here . Thanks –  Mar 02 '18 at 14:27

1 Answers1

1

The title announces the topic of resolving inequalities, while the OP is largely devoted to the topic of their compilation. Both topics are interesting, but accents must be placed.

INEQUALITIES COMPILATION

  1. The consideration can be supplemented by references to inequalities already compiled by the author. For example, this one.

  2. An integral approach should not be applied to functions of several variables, because $x, y$ and $z$ in general are the independent variables. Really, this leads to ODE.
    Suppose that one is trying to obtain a maximum of $$g(x,y) = f(x) + f(y), \text{ where } xy = 1 \wedge (x,y) = (0,\infty)^2,$$ and have got the equation $$g\left(x,\dfrac1x\right) = f(x) + f\left(\dfrac1x\right) = \dfrac{x^{10}}{x^{11}+1} + \dfrac {x}{x^{11}+1}.\tag1$$ Let the antiderivative $F(x)$ is given. How to use it to integrate both of terms $(1)?$

  3. The idea of integration can be used to generate functions with the required properties (for example, slow and fast ascending intervals).

  4. The integration operation summarizes the data on the adjacent intervals and contains the constant. So RHS of the inequality should be considered separately, and obtained inequalities should have the clear solution.

INEQUALITIES SOLUTION

  1. The maximum function problem.
    Let us analyze one of the proposed inequations in the form of $G(x,y) \le 1,\ $ where $$G(x,y) = g(1/x) + g(xy) + x - y,\quad y+0.1 \ge x\ge y \ge 1,\tag2$$ $$g(t) = f(1/t) = \dfrac{t}{1+t^{11}},\tag3$$ $$g'(t) =\dfrac{1-10t^{11}}{(1+t^{11})^2}.\tag4$$ The inequality $(2)$ can be solved as the maximum function problem with the nesessary conditions of extremum $G'_x = G'_y = 0,$ or $$\begin{cases} -1/x^2g'(1/x) + yg'(xy) + 1 = 0\\[4pt] xg'(xy) - 1 = 0, \end{cases}$$

  2. The algebraic system solution. $$\begin{cases} -1/xg'(1/x) + y + x = 0\\[4pt] x\dfrac{1-10(xy)^{11}}{(1+(xy)^{11})^2} = 1, \end{cases}$$ $$\begin{cases} y = 1/xg'(1/x) -x\\[4pt] x^{22}y^{22}+2(5x+1)x^{11}y^{11} - x+1 = 0, \end{cases}$$ $$\begin{cases} y = x^{10}\dfrac{x^{11}-10}{(1+x^{11})^2} - x\\[4pt] \left[ \genfrac{}{}{0}{0} {y = \dfrac1x\sqrt[11]{-5x-1 - \sqrt{(25x+11)x}}} {y = \dfrac1x\sqrt[11]{-5x-1 + \sqrt{(25x+11)x}}.} \right. \end{cases}$$ The graphic solutions for the "negative" branch are in the Plot1.
    Plot1
    The graphic solutions for the "positive" branch are in the Plot2.
    Plot2
    There are not solutions with $y> -0.704459.$ So the highest value of $LSH(2)$ are in the boundary of the domain definition.

  3. The boundary checking.
    The counterexample was found on the upper bound: $$max\left\{\dfrac{x^{10}}{1 + x^{11}} + \dfrac{xy}{1 + (xy)^{11}} + x - y|y + 0.1 = x>=y>=1\right\}≈1.05864$$ $$ at (x, y) = (1.1, 1).$$

  4. Conclusion
    There are some ways to fix the problem. The most interesting one seems the increasing of RSH constant and inequality domain definition.

USING OF THE MICHAEL ROZENBERG INEQUALITY FEATURES IN THE SOLUTION

  1. Actual conditions can sometimes be used by differentiation.
    Let us consider the inequality $\Phi(x,y,z), \le3$ where $$\Phi(x,y,z) = g(x) + g(y) + g(z), \quad xyz = 1 \wedge (x,y,z)\in(0,\infty)^3\tag5$$ and $g(x)$ is given by $(3).$

  2. The elimination of variable $z = \dfrac{1}{xy}$ allows to solve the inequality as the maximum function problem for $$G(x,y) = g(x) + g(y) + g\left(\dfrac1{xy}\right)$$ with the nesessary extremum conditions $G'_x = G'_y = 0,$ or $$\begin{cases} g'(x) - \dfrac{1}{x^2y}g'\left(\dfrac1{xy}\right) = 0\\[4pt] g'(y) - \dfrac{1}{xy^2}g'\left(\dfrac1{xy}\right) = 0, \end{cases}$$ or $$\dfrac1xg'(x) = \dfrac1yg'(y) = \dfrac1{xy}g'\left(\dfrac1{xy}\right).\tag6$$

  3. Let $$h(t) = \dfrac1tg'(t),$$ then the system $(6)$ can be presented in the form $$h(x) = h(y) = h(z).\tag7$$ and this is a nice feature of a task.

4 The further solution is based on the searching of situations where the pair of variables $x,y,z$ belong to one monotonical branch of $h(t).$
The full solution of the inequality is there.

Yuri Negometyanov
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