Prove or disprove: $1+\frac{\ln^c(ax)-\ln^ca}{\ln^c(2a)-\ln^ca}-x\leq 0$, where $12$, and $c$ is chosen so that $f'(2)=0$

Question

Define:

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

where $1< x$ and $a>2$. Assume further that the parameter $c$ is chosen so that $f'(2)=0$. The derivation of $c$ involves the Lambert's function.

Claim : $$f(x)\leq 0$$

My attempt:

We have :

$$f'(x)=\frac{c(\ln(ax))^{c-1}}{x\left(\ln^{c}\left(2a\right)-\ln^{c}\left(a\right)\right)}-1$$

We substitute $x=\frac{1}{y^{c-1}a}$

The inequality have the form :

$$\ln(u)u=p$$

Wich is just the Lambert's function .See the solution in this link . I cannot proceed further .

How to (dis)prove the first inequality ?

Thanks in advance

How do you compute $c$ from $f'(2)=0$ ? I do not see Lambert function as a solution. But you know, at my age, I am dumb. Cheers :-) — Claude Leibovici, Jul 02 '21 at 12:01
The question would me much clearer if you separated the definition of $f$ (with its parameters $a,c$, where the latter is determined implicitly) from the statement of the problem. — Michał Miśkiewicz, Jul 02 '21 at 12:09
Well, it still wasn't clear to me, so I suggested an edit explaining the role of $c$. If you don't like it, feel free to undo it. — Michał Miśkiewicz, Jul 02 '21 at 12:21
You say you know this equation holds $x < 1$, but then you use that equation to calculate $f'(2)$? Does the equation hold for more than $x < 1$? — Paul Sinclair, Jul 03 '21 at 02:44
@PaulSinclair Hum I suspect you don't read carefully the definition .I say $x>1$ otherwise it's totally wrong . — Miss and Mister cassoulet char, Jul 03 '21 at 11:26

River Li · Accepted Answer · 2021-07-25T23:28:32.887

1

Sketch of a proof:

From $f'(2) = 0$, we have $$\left(\frac{\ln a}{\ln (2a)}\right)^c = 1 - \frac{c}{2\ln (2a)}.$$ Let $$g(c) = \left(\frac{\ln a}{\ln (2a)}\right)^c - 1 + \frac{c}{2\ln (2a)}.$$ We have \begin{align*} g'(c) &= \left(\frac{\ln a}{\ln (2a)}\right)^c\ln \frac{\ln a}{\ln (2a)} + \frac{1}{2\ln (2a)},\\ g''(c) &= \left(\frac{\ln a}{\ln (2a)}\right)^c \left(\ln \frac{\ln a}{\ln (2a)}\right)^2 > 0. \end{align*} Clearly $g(0) = 0$, $g'(0) < 0$, $g(1) = \frac{1 - 2\ln 2}{2\ln (2a)} < 0$ and $g(\infty) = \infty$.

Fact 1: $g(1 + \ln(2a)) > 0$.

Fact 2: $1 < c < 1 + \ln (2a)$.

Now, we have $$f''(x) = \frac{c\ln^c (ax)}{[\ln^c (2a) - \ln^c a]x^2 \ln^2 (ax)}[c - 1 - \ln (ax)].$$ Let $x_0 = \frac{1}{a}\mathrm{e}^{c - 1}$. We have $f''(x_0)= 0$, $f''(x) > 0$ on $(0, x_0)$, and $f''(x) < 0$ on $(x_0, \infty)$. Also, by Fact 2, we have $x_0 \in (0, 2)$.

Since $f(2) = 0$ and $f'(2) = 0$, noting that $f''(x) \le 0$ on $[x_0, \infty)$, we have $f(x) \le f(2) = 0$ for all $x \in [x_0, \infty)$.

Since $f(1) = 0$ and $f(x_0) \le 0$, noting that $f''(x) \ge 0$ on $(0, x_0]$, we have $f(x) \le \max\{f(1), f(x_0)\} = 0$ for all $x \in (1, x_0]$. (Note: If $g(x)$ is a convex function on $[a, b]$, then $g(x) \le \max\{g(a), g(b)\}$.)

We are done.

edited Jul 25 '21 at 23:28

answered Jul 25 '21 at 15:37

River Li

37,323

Nice to see you here ! Let me the time to check it properly .As you know I have always some questions but just one here : Do we have some convergence when $a\to \infty$? – Miss and Mister cassoulet char Jul 27 '21 at 10:28
@ErikSatie Why do you need to consider $a \to \infty$? – River Li Jul 27 '21 at 10:56
Because I try to show the inequality https://math.stackexchange.com/questions/2198645/prove-that-if-ab-1-then-forall-n-in-mathbbn-a2bn-b2a?noredirect=1&lq=1 . With the form $((1-x)^{-(2x)^n}-1)((x)^{-(2(1-x))^n}-1)\geq 1$ .So I need some accuracy . – Miss and Mister cassoulet char Jul 27 '21 at 11:00
@ErikSatie Some convergence of what? – River Li Jul 27 '21 at 11:04
Someting like $\lim_{a,c\to\infty}f(x)=0$ If you see what I mean with $x>1$ – Miss and Mister cassoulet char Jul 27 '21 at 11:41
@ErikSatie $f(x)$ is concave on $(x_0, \infty)$ and $f(2)$ is global maximum, so $\lim_{a\to \infty} f(x)\ne 0$. – River Li Jul 27 '21 at 12:09
Fine ok thanks ! – Miss and Mister cassoulet char Jul 27 '21 at 13:14

Prove or disprove: $1+\frac{\ln^c(ax)-\ln^ca}{\ln^c(2a)-\ln^ca}-x\leq 0$, where $12$, and $c$ is chosen so that $f'(2)=0$

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