Prove that $f(x_0)>\frac{2}{3}$

Question

It's a problem found with the help of Geogebra.

Let $0<x$ be a real number then define the function:

$$f(x)=\Big(\frac{x}{x+1}\Big)^{\Gamma(x)}$$ Then let $x_0$ be the maximum of the function on $(0,\infty)$ and then prove that:

$$f(x_0)>\frac{2}{3}$$

See here to compare

Well to solve it I have tried logically the use of derivative we have:

$$f'(x)=\Big(\frac{x}{x+1}\Big)^{\Gamma(x)} \Bigg(\frac{(x + 1) \Big(\frac{1}{(x + 1)} - \frac{x}{(x + 1)^2}\Big) Γ(x)}{x} + \log\Big(\frac{x}{x + 1}\Big) Γ(x) \psi^{(0)} (x)\Bigg)$$

Where we have the $n^{th}$ derivative of the digamma function.

I think that this derivative is not really useful only theoretically, but we can use the Newton's method numerically .

I have tried some inequality on the this wiki page notably an inquality due to Kečkić and Vasić without success.

On the other hand the problem with Taylor series is : we get a lot of constant as Euler-Mascheroni constant wich needs to be evaluate with an series or something like that. So it's a little bit make problem on another problem.

Maybe spline cubic is the way I don't know...

Finally taking the logarithm on both side the derivative is a little bit less tedious.See here

Well if you have an issue thanks in advance ...

Answer 1

Hint

Try to expand $f$ at the first order around $2$ based on

$g(x) = \frac{x}{x+1} = \frac{2}{3}(1+h/6) +o(h^2)$ where $x=2+h$ and $\Gamma(2+h)=1+(1-\gamma)h+o(h^2)$ where $\gamma$ is the Euler Mascheroni constant.

Therefore $$\begin{aligned} \ln f(2+h) &= (1+(1-\gamma)h+o(h^2))(\ln(2/3) + h/6 + o(h^2))\\ &=\ln(2/3) + ((1-\gamma)\ln(2/3) + 1/6)h +o(h^2) \end{aligned} $$ proving that $f$ takes around $2$ values larger than $2/3$ as $(1-\gamma)\ln(2/3) + 1/6 \neq 0$.

Answer 2

Almost the same as in comments and answers.

Since $x>0$, maximizing $$f(x)=\Big(\frac{x}{x+1}\Big)^{\Gamma(x)}$$ is the same as maximizing $$g(x)=\Gamma(x) \log\Big(\frac{x}{x+1}\Big)$$ for which $$\frac{g'(x)}{g(x)}=\Gamma (x) \left(\frac{1}{x(x+1)}+\log \left(\frac{x}{x+1}\right) \psi (x)\right)$$ and, as already said, the quantity inside parentheses cancels close to $x=2$. Using one single iteration of Newton, Halley, Householder and higher order iterative methods of the same class, we obtain totally explicit expressions of $x_0$ corresponding to the maximum of $f(x)$. Since the formulae can be quite long, only their decimal representation will be given as a function of $n$ (the order of the method). $$\left( \begin{array}{ccc} n & x_0^{(n)} & \text{method} \\ 2 & 1.985579580 & \text{Newton}\\ 3 & 1.985734229 & \text{Halley}\\ 4 & 1.985733904 & \text{Householder}\\ \cdots & \cdots & \text{no name}\\ \infty & 1.985733904 & \end{array} \right)$$ So, $$x_0^{(2)}=2+\frac{36 (\gamma -1) \log \left(\frac{3}{2}\right)-66}{35+6 \gamma +6 \left(\pi^2-6\right) \log \left(\frac{3}{2}\right)}$$ seems to be a sufficient approximation.

$$f(x_0^{(2)})=\Big(\frac{x_0^{(2)}}{x_0^{(2)}+1}\Big)^{\Gamma(x_0^{(2)})}\approx 0.6666893243$$ Notice that a full optimization gives a maximum of $0.6666893270$ for $x=1.985733903$.