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

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

Let $$0 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 ...

• Not sure if this will help but I will still put it out there. Notice that $f(2) = 2/3$ and you can show that the derivative at $x = 2$ is negative (I think it is just less than $0$), then using continuity you can claim at that for some $\epsilon > 0$, the value of $f(2 - \epsilon)$ would be $2/3 + \delta$ for $\delta > 0$. It would also hold for the maximum value then. Jul 30, 2020 at 12:37
• @sudeep5221 That observation settles it pretty quickly. And verifying that $$(1 - \gamma)\log \frac{3}{2} \neq \frac{1}{6}$$ doesn't require high precision in the computation of $\gamma$ and $\log \frac{3}{2}$. Please consider expanding your comment into an answer. Jul 30, 2020 at 13:20

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 $$(1-\gamma)\ln(2/3) + 1/6 \neq 0$$.

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 $$x=1.985733903$$.