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my question is about an inequality from Michael Rozenberg

See here for the initial inequality

My trying :

- Prove that powers of any fixed prime $p$ contain arbitrarily many consecutive equal digits.
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We make the following substitution with $b\geq a $ et $c\geq a$:

$A=a$

$AB=b$

$AC=c$

After simplification we get :

$$\frac{1}{13+5B^2}+\frac{B^3}{13B^2+5C^2}+\frac{C^3}{13C^2+5}\geq \frac{1+B+C}{18}$$

This makes one study a much more general inequality than the initial one and apply the weighted Karamata inequality . All the subtlety lies in the fact that one does not take the side Right of inequality but work only on the left side.

Thus the more general inequality take this form:

$$\frac{1}{13+5x^2}+\frac{x^3}{13x^2+5y^2}+\frac{y^3}{13y^2+5}\geq \frac{1}{13+5z^2}+\frac{z^3}{13z^2+5u^2}+\frac{u^3}{13u^2+5}$$

With $x$,$y$,$u$,$z$ positiv real number and the condition $ y \geq u\geq 1$ and $z\geq 1$ and $xu \geq yz$:

So we apply the theorem 3 from the following link

to the funtion $g(v)=\frac{v^3}{13v^2+5}$ wich respect the condition of the theorem 3(convex+increase)

With :

$x_3=y$

$x_2=\frac{x}{y}$

$x_1=\frac{1}{x}$

$y_3=u$

$y_2=\frac{z}{u}$

$y_1=\frac{1}{z}$

$p_3=1$

$p_2=y$

$p_1=x$

And:

$x_3\geq x_2\geq x_1$ or $x_2\geq x_3\geq x_1$ and $y_3\geq y_2\geq y_1$ or

$y_2\geq y_3\geq y_1$

We have :

$p_3x_3\geq p_3y_3$ wich correspond to $y\geq u$

$p_2x_2\geq p_2y_2$ wich correspond to $xu\geq yz$

$p_1x_1\geq p_1x_1$ wich correspond to $1\geq 1$

Therefore, the majorisation is acquired whatever the order of $x_i$ et $y_i$

So we find:

$$\frac{1}{13+5x^2}+\frac{x^3}{13x^2+5y^2}+\frac{y^3}{13y^2+5}\geq \frac{x}{z}\frac{1}{13+5z^2}+\frac{y}{u}\frac{z^3}{13z^2+5u^2}+\frac{u^3}{13u^2+5}$$

Moreover we have $\frac{y}{u}\geq 1$ and $\frac{x}{z}\geq 1$ we get :

$$\frac{1}{13+5z^2}+\frac{z^3}{13z^2+5u^2}+\frac{u^3}{13u^2+5}\leq \frac{x}{z}\frac{1}{13+5z^2}+\frac{y}{u}\frac{z^3}{13z^2+5u^2}+\frac{u^3}{13u^2+5}$$

So we have the inequality of the beginning :

$$\frac{1}{13+5x^2}+\frac{x^3}{13x^2+5y^2}+\frac{y^3}{13y^2+5}\geq \frac{1}{13+5z^2}+\frac{z^3}{13z^2+5u^2}+\frac{u^3}{13u^2+5}$$

Now we make the following substitution :

$z=u=\frac{x+y}{2}$

We have :

$$\frac{1}{13+5(\frac{(x+y)}{2})^2}+\frac{(\frac{(x+y)}{2})^3}{13(\frac{(x+y)}{2})^2+5(\frac{(x+y)}{2})^2})+\frac{(\frac{(x+y)}{2})^3}{13(\frac{(x+y)}{2})^2+5}\geq \frac{1+(x+y)}{18}$$

We prove this for $x\geq 1$,$y\geq 1$ positive real number :

$$\frac{1}{13+5(\frac{(x+y)}{2})^2}+\frac{(\frac{(x+y)}{2})^3}{13(\frac{(x+y)}{2})^2+5(\frac{(x+y)}{2})^2}+\frac{(\frac{(x+y)}{2})^3}{13(\frac{(x+y)}{2})^2+5}\geq \frac{1+(x+y)}{18}$$

We can simplify it :

$$\frac{1}{13+5(\frac{(x+y)}{2})^2}+\frac{(\frac{(x+y)}{2})^3}{13(\frac{(x+y)}{2})^2+5}\geq \frac{1+\frac{(x+y)}{2}}{18}$$

We make the substitution $k=\frac{x+y}{2}$

$$\frac{1}{13+5(k)^2}+\frac{(k)^3}{13(k)^2+5}\geq \frac{1+k}{18}$$

We can rewrite this in an other form :

$$\frac{5(k-1)^2(k+1)(5k²-8k+5)}{18(13k^2+5)(13+5k^2)}\geq 0$$

So my question is What are the cases I have forgotten to fully demonstrate the initial inequality ?

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- A problem with 26 distinct positive integers
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Theorem 3 from the link requiers both $(x_1,x_2,x_3)$ and $(y_1,y_2,y_3)$ be decreasing $n$-tuples, whereas you state the opposite: $x_3\geq x_2\geq x_1$ or $x_2\geq x_3\geq x_1$ and $y_3\geq y_2\geq y_1$ or $y_2\geq y_3\geq y_1$. Anyway, this fails when $x=z=1$ and $y=u=2$, because then $x_1=1> 1/2=x_2$. On the other hand, it requires a weaker $p_i$-related majorization conditions than those which your prove. Nevertheless, a computer calculational evidence suggests that “the inequality of the beginning” is true.

Next, the substitution $z=u=\frac{x+y}2$ is valid iff $x\ge y$ (which corresponds to the case $b\ge c$ in the initial inequality), and the remaining case may be a main problem of your approach, because the initial inequality is cyclic but not symmetric.

The rest of your proof is OK.

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