Maximum of $\frac{x_1(1-x_1)x_2(1-x_2)...x_n(1-x_n)}{1-(1-x_1x_2...x_{n-1})x_n}$ for $x_i \in (0, 1), \forall i$

Question

With reference to this irrationality proof of $\zeta(3)$, I need to find the maximum of $$f(x_1,x_2,...,x_n)= \frac{x_1(1-x_1)x_2(1-x_2)...x_n(1-x_n)}{1-(1-x_1x_2...x_{n-1})x_n}$$ where $0<x_i<1$ for all $1\leq i\leq n$.

I tried to find an upper bound of $f$ which is not its maximum value as follows: $$1-(1-x_1x_2...x_{n-1})x_n=1-x_n+x_1x_2...x_{n}\geq 2\sqrt{1-x_n}\sqrt{x_1x_2...x_{n}} $$ where the last inequality follows from AM-GM inequality since $1-x_n>0$ and $x_i>0$ for all $1\leq i\leq n$. So we get $$f(x_1,x_2,...,x_n)\leq \frac{\sqrt{x_1}(1-x_1)\sqrt{x_2}(1-x_2)...\sqrt{x_{n-1}}(1-x_{n-1})\sqrt{x_{n}(1-x_{n})}}{2}$$ Since for all $1\leq i\leq n-1$, $\max_{0<x_i<1}{\sqrt{x_i}(1-x_i)}= \frac{1}{\sqrt{3}}(1-\frac{1}{3})$ and $\max_{0<x_n<1}{\sqrt{x_{n}(1-x_{n})}}=\frac{1}{2}$, so we get $$f(x_1,x_2,...,x_n)\leq \frac{(\frac{2}{3\sqrt{3}})^{n-1}}{4}$$ So for $n=3$, $$\frac{x_1(1-x_1)x_2(1-x_2)x_3(1-x_3)}{1-(1-x_1x_2)x_3}\leq \frac{(\frac{2}{3\sqrt{3}})^{2}}{4}\approx 0.037037037$$
In this irrationality proof of $\zeta(3)$ we have $$\frac{x_1(1-x_1)x_2(1-x_2)x_3(1-x_3)}{1-(1-x_1x_2)x_3}\leq (\sqrt{2}-1)^4\approx 0.02943725$$ So the above method does not give the maximum. Any help will be highly appreciated. Thank you!

Edit What is the maximum value of $f(x_1,x_2,...,x_{11})$?

Answer 1

This is a partial answer reducing the problem to two variables, with no differential calculus. Next one variable can be eliminated by studying the partial derivative.

First of all the function is is positive for $0<x_k<1,$ $1\le k\le n$ and vanishes on the boundary $x_k=0$ or $x_k=1$ for some $1\le k\le n.$ Therefore it attains the maximal value inside the region.

Let $n\ge 3.$ Assume the maximum is attained at $(a_1,a_2,\ldots, a_n).$ The expression is of the form $$A_n{a_1a_2[1+a_1a_2-(a_1+a_2)]\over B_n+C_na_1a_2}$$ where $A_n,B_n,C_n>0$ are independent of $a_1$ and $a_2.$ Given the product $a_1a_2$ the maximal value is attained when $a_1+a_2$ is the smallest possible, i.e. $a_1=a_2.$ This reasoning implies that $a_1=a_2=\ldots=a_{n-1}.$ Therefore we are down to the function in two variables $x:=x_1=\ldots =x_{n-1}$ and $y:=x_n$ of the form $$g(x,y):=(1-x)^{n-1}{x^{n-1}y(1-y)\over 1-(1-x^{n-1})y}$$
The equation $\partial g/\partial y=0$ leads to $$x^{n-1}y^2-2y+1=0$$ Eliminating either $y$ or $x$ we obtain a one variable function of either $x$ or $y.$

Answer 2

Assuming that the minimum is located in a interior point to $0\le x_i\le 1, i = 1,\cdots,n$ and calling $p_n(x) = \prod_{i=1}^nx_i$ and $q_n(x) = \prod_{i=1}^n(1-x_i)$ we have

$$ f_n(x) = \frac{p_n(x)q_n(x)}{1-x_n+p_n(x)} $$

so having in mind that $\frac{d}{d x_k}p_n(x) = \frac{p_n(x)}{x_k}$ and $\frac{d}{d x_k}q_n(x) =-\frac{q_n(x)}{1-x_k}$ we have for $k\ne n$

$$ \frac{d}{d x_k}f_n(x) = q_n(x)p_n(x)\frac{x_k p_n(x)+x_n-2x_k(x_n-1)-1}{x_k(x_k-1)(1-x_n+p_n(x))^2}=0 $$

and for $k=n$ analogously

$$ \frac{d}{d x_n}f_n(x) =\frac{q_n(x)p_n(x)(x_np_n(x) -(x_n-1)^2)}{x_n(x_n-1)(1-x_n+p_n(x))^2}=0 $$

and eliminating $p_n(x)$ we arrive at

$$ x_k = \frac{x_n}{x_n+1} $$

The general expression for $f_n(x)$ can be written as

$$ f_n(x) = f_n(x_n) = \frac{(1-x_n) x_n^n \left(1-\frac{x_n}{x_n+1}\right)^{n-1}}{(x_n+1)^{n-1} \left(\frac{x_n^n}{(x_n+1)^{n-1}}-x_n+1\right)} $$

The maximum can be obtained with a procedure to obtain the roots of a polynomial by solving

$$ f_n(x_n)'= 0 $$

NOTE

Calculating for $n=3$ we obtained

$$ f_3 = \frac{1-\frac{1}{\sqrt{2}}}{2 \sqrt{2} \left(\frac{1}{2}+\frac{1}{\sqrt{2}}\right) \left(1+\frac{1}{\sqrt{2}}\right)^2}\approx 0.02943725152285942 $$

for $x_3 = \frac{1}{\sqrt{2}}$

For $n = 4$ we obtained

$$ f_4 \approx 0.008964838545691446, \ \ \ x_4 \approx 0.7759185953243913 $$

Attached a MATHEMATICA script to calculate for $n$ and shown in B

Clear[p, q, Fn]
B = {};
p[n_, x_] := x Product[x/(1 + x), {i, 1, n - 1}]
q[n_, x_] := (1 - x ) Product[1 - x/(1 + x), {i, 1, n - 1}]
Fn[n_, x_] := p[n, x] q[n, x]/(1 - x + p[n, x])
For[k = 2, k <= 10, k++,
  Fkmax = -10^10;
  sols = Solve[D[Fn[k, x], x] == 0, x, Reals];
  nsols = Length[sols];
  For[j = 1, j <= nsols, j++,
   xk = x /. sols[[j]];
   If[0 <= xk <= 1, Fk = Fn[k, x] /. sols[[j]], Fk = -10^10];
   If[Fk >= Fkmax, Fkmax = Fk; xkmax = xk]];
  AppendTo[B, {Fkmax, xkmax}]];

B // N

Answer 3

Let $$u := x_1x_2 \cdots x_{n-1} \in (0, 1).$$

By AM-GM, we have \begin{align*} \frac{x_n(1 - x_n)}{1 - (1 - u)x_n} &= \frac{1 + u}{(1 - u)^2} - \left[\frac{1 - (1 - u)x_n}{(1 - u)^2} + \frac{u}{(1 - u)^2}\cdot \frac{1}{1 - (1 - u)x_n}\right]\\ &\le \frac{1 + u}{(1 - u)^2} - 2\sqrt{\frac{1 - (1 - u)x_n}{(1 - u)^2} \cdot \frac{u}{(1 - u)^2}\cdot \frac{1}{1 - (1 - u)x_n}}\\ &= \frac{1 + u}{(1 - u)^2} - \frac{2\sqrt u}{(1 - u)^2}\\ &= \frac{1}{(1 + \sqrt u)^2} \tag{1} \end{align*} with equality if $x_n = \frac{1}{1 + \sqrt u}$.

By AM-GM, we have \begin{align*} &(1 - x_1)(1 - x_2)\cdots (1 - x_{n-1})\\ \le{}& \frac{1}{(n - 1)^{n - 1}}(1 - x_1 + 1 - x_2 + \cdots + 1 - x_{n-1})^{n-1}\\ ={}& \frac{1}{(n - 1)^{n - 1}}\Big[n - 1 - (x_1 + x_2 + \cdots + x_{n-1})\Big]^{n-1}\\ \le{}& \frac{1}{(n - 1)^{n - 1}} \Big[n - 1 - (n - 1)\sqrt[n-1]{x_1x_2 \cdot x_{n-1}}\Big]^{n-1}\\ ={}& \left(1 - \sqrt[n-1]{u}\right)^{n-1}\tag{2} \end{align*} with equality if $x_1 = x_2 = \cdots = x_{n - 1}$.

From (1) and (2), we have $$f(x_1, x_2, \cdots, x_n) \le g(u) := u \cdot \frac{1}{(1 + \sqrt u)^2} \cdot \left(1 - \sqrt[n-1]{u}\right)^{n-1} \tag{3}$$ with equality if $x_1 = x_2 = \cdots = x_{n-1}$ and $x_n = \frac{1}{1 + \sqrt{x_1x_2 \cdots x_{n-1}}}$.

Thus, equivalently, we turn to find the maximum of $g(u)$ subject to $u \in (0, 1)$.
(Note: Since $g(0)= g(1) = 0$, this optimization problem is well-defined.)

Letting $v = u^{\frac{1}{2(n-1)}} \in (0, 1)$, equivalently, we turn to find the maximum of $$h(v) := \frac{v^{2n - 2}(1 - v^2)^{n-1}}{(1 + v^{n - 1})^2}$$ subject to $v \in (0, 1)$. Since $h(0) = h(1) = 0$, this optimization problem is well-defined. Also, the maximum occurs at some $v \in (0, 1)$ with $h'(v) = 0$.

We have $$h'(v) = - \frac{2(n - 1)v^{2n-2}(1 - v^2)^{n - 2}(v^{n + 1} + 2v^2 - 1)}{v(1 + v^{n-1})^3}.$$ Let $v_0 \in (0, 1)$ be the unique real root of $v^{n + 1} + 2v^2 - 1 = 0$ on $(0, 1)$. Note that $v_0$ is the unique real root of $h'(v) = 0$ on $(0, 1)$. Thus, the maximum of $h$ is $h(v_0)$.

Thus, the maximum of $f(x_1, x_2, \cdots, x_n)$ is equal to $h(v_0)$ when $x_1 = x_2 = \cdots = x_{n-1} = v_0^2$ and $x_n = \frac{1}{1 + v_0^{n-1}}$, where $v_0 \in (0, 1)$ be the unique real root of $v^{n + 1} + 2v^2 - 1 = 0$ on $(0, 1)$.