0
$\begingroup$

I came up with the following inequality, but I am not able to prove it, despite the plots suggesting that the following inequality is true: $$-\gamma x^2 ( x^{k/\gamma}-1 ) -x^2 + \alpha x - \alpha + 1 \leq 0,\qquad\ 2 \leq x \leq n-1.$$

Let me explain this inequality in more details since I used some notations to simplify it.

$n \geq 3$ is an integer, and so is $2 \leq x \leq n-1$. For the purpose of the proof, it might be easier to consider that $x$ is a real number.

$\alpha = \alpha(n) = \left \lceil \frac{n}{\log n} \right \rceil$ depends only on $n$.

$\gamma = \gamma(n, x)$ is actually a function of $n$ and $x$ defined by $$\gamma(n,x) = n \cdot \frac{x + W(-xe^{-x})}{x}$$ where $W(z)$ is the Lambert's $W$ function.

$k = k(n,x)$ is also a function of $n$ and $x$ defined by $$k(n,x) = \frac{\log \left( \frac{\log(n) - 1}{\log n} \right)}{\log \left( \frac{n - x}{n} \right)}.$$

Now that my notations are fixed, let $n \geq 3$ be fixed. How can I prove that $$-\gamma x^2 ( x^{k/\gamma}-1 ) -x^2 + \alpha x - \alpha + 1 \leq 0,\ 2 \leq x \leq n-1?$$

Here's a plot depicting what's happening: Plot

I tried to derive the function $f(x) = -\gamma x^2 ( x^{k/\gamma}-1 ) -x^2 + \alpha x - \alpha + 1$, but since on the plots $f$ is not strictly decreasing, that was not very useful (also the derivative is ugly).

So do you have any hints/idea on what I could try to prove this?

$\endgroup$
  • $\begingroup$ Isn't $W(-xe^{-x}) = -x$ since $W$ is the inverse of $z \mapsto ze^{z}?$ In which case your $\gamma$ is identically $0,$ and stuff like $x^{k/\gamma}$ is not well defined. $\endgroup$ – stochasticboy321 Feb 26 '18 at 21:48
  • 1
    $\begingroup$ @stochasticboy321 No, this identity is true for $x \geq-1$ and unfortunately, I have $x \leq-2$. $\gamma$ converges to $n$, and $x^{k/\gamma} $ converges to $0$. $\endgroup$ – Laurent Hayez Feb 26 '18 at 22:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.