, 2022

Proof. The function $F$, viewed as a continuous function with respect to $x_1, x_2, \cdots, x_{11}$, has a maximum value on the bounded closed set
$$\Omega = \{(x_1, x_2, \cdots, x_{11}) \in (\mathbb{R}_{\ge 0})^{11} \mid x_1 + x_2 + \cdots + x_{11} = 1\}.$$
Assume that $F$ has a maximum value at $(a_1, a_2, \cdots, a_{11}) \in \Omega$. It is clear that $F > 0$ at this point.

(1) $a_3 = a_5 = a_7 = a_9 = a_{11} = 0$.
If $a_3 > 0$, let $a'_3 = 0$, $a'_4 = a_3 + a_4$, and let $a'_i = a_i$ for other $i$. This maintains the sum. We prove that $F(a_1, a_2, \cdots, a_{11}) < F(a'_1, a'_2, \cdots, a'_{11})$, which is equivalent to
$$(a_2 + a_3 + a_4)(a_3 + a_4)(a_4 + a_5 + a_6) < (a'_2 + a'_3 + a'_4)(a'_3 + a'_4)(a'_4 + a'_5 + a'_6),$$
and since $a_3 + a_4 = a'_3 + a'_4$ and $a_4 < a'_4$, this inequality holds. This contradicts the fact that $F$ has a maximum value at $(a_1, a_2, \cdots, a_{11})$. Therefore, $a_3 = 0$. Similarly, we can prove that $a_5 = a_7 = a_9 = a_{11} = 0$.

Now, let $x_3 = x_5 = x_7 = x_9 = x_{11} = 0$ in the expression for $F$. We will now view $F$ as a function of $x_1, x_2, x_4, \cdots, x_{10}$,
$$F = (x_1 + x_2)(x_2 + x_4)x_4(x_4 + x_6)x_6(x_6 + x_8)x_8(x_8 + x_{10})x_{10}(x_{10} + x_1)x_1.$$
Then, $F$ has a maximum value at $(a_1, a_2, a_4, a_6, a_8, a_{10})$. It is clear that $a_1, a_4, a_6, a_8, a_{10} > 0$.

(2) $a_2 = 0$.
By contradiction, suppose $a_2 > 0$. If $a_1 \le a_4$, let $a'_1 = a_1 + a_2$, $a'_2 = 0$, and let $a'_i = a_i$ for other $i$. This maintains the sum. We prove that $F(a_1, a_2, a_4, \cdots, a_{10}) < F(a'_1, a'_2, a'_4, \cdots, a'_{10})$.
$$\Leftrightarrow a_1(a_{10} + a_1)(a_1 + a_2)(a_2 + a_4) < a'_1(a'_{10} + a'_1)(a'_1 + a'_2)(a'_2 + a'_4).$$
Since $a_1 < a'_1$ and $a_{10} + a_1 < a'_{10} + a'_1$, we only need to prove:
$$a_1(a_1 + a_2)(a_2 + a_4) \le a'_1(a'_1 + a'_2)(a'_2 + a'_4) = (a_1 + a_2)(a_1 + a_2)a_4,$$
$$\Leftrightarrow a_1(a_2 + a_4) \le a_4(a_1 + a_2), \quad \Leftrightarrow a_1a_2 \le a_4a_2,$$
which holds. If $a_1 \ge a_4$, let $a'_4 = a_4 + a_2$, $a'_2 = 0$, and for other $i$, let $a'_i = a_i$. We also have:
$$F(a_1, a_2, a_4, \cdots, a_{10}) < F(a'_1, a'_2, a'_4, \cdots, a'_{10}).$$
This contradicts the fact that $F$ reaches its maximum value at $(a_1, a_2, a_4, a_6, a_8, a_{10})$. Therefore, $a_2 = 0$.

Now let $x_2 = 0$ in the analytic expression of $F$, and regard $F$ as a function of $x_1, x_4, x_6, x_8, x_{10}$,
$$F = x_4^2(x_4 + x_6)x_6(x_6 + x_8)x_8(x_8 + x_{10})x_{10}(x_{10} + x_1)x_1^2.$$
Then $F$ reaches its maximum value at $(a_1, a_4, a_6, a_8, a_{10})$.

(3) $a_1 = a_4, a_6 = a_{10}$.
Let $a'_1 = a'_4 = \frac{1}{2}(a_1 + a_4)$, $a'_6 = a'_{10} = \frac{1}{2}(a_6 + a_{10})$, $a'_8 = a_8$, keeping the sum unchanged. By the inequality of arithmetic and geometric means,
$$\text{a 1 2 a 4 2 a' 1 2 a' 4 2, (a 4 + a 6)(a 10 + a 1) (a' 4 + a' 6)(a' 10 + a' 1),}$$
$$a_6 a_{10} \le a'_6 a'_{10}, \quad (a_6 + a_8)(a_8 + a_{10}) \le (a'_6 + a'_8)(a'_{8} + a'_{10}).$$
Also, $a_8 = a'_8$. Multiplying these equations gives:
$$F(a_1, a_4, \dots, a_{10}) \le F(a'_1, a'_4, \dots, a'_{10}).$$
Since $F$ reaches its maximum value at $(a_1, a_4, \dots, a_{10})$, the equality holds in the above inequality. By the conditions for equality in the inequality of arithmetic and geometric means, we have $a_1 = a_4, a_6 = a_{10}$.

In the analytic expression of $F$, replace $x_4$ with $x_1$, and $x_{10}$ with $x_6$. Now regard $F$ as a function of $x_1, x_6, x_8$,
$$F = x_1^4 x_6^2 x_8 (x_1 + x_6)^2 (x_6 + x_8)^2,$$
and $2x_1 + 2x_6 + x_8 = 1$.

(4) Next, we will solve the system of equations (this system of equations is obtained by matching coefficients in the inequality of arithmetic and geometric means, which will be used in (5)):
$$\frac{2}{u} + \frac{1}{u+v} = \frac{1}{v} + \frac{1}{u+v} + \frac{1}{v+1} = 1 + \frac{2}{v+1}, \quad u, v > 0.$$
We will prove that there is a unique positive real solution $u, v$, and that $v < 1$.
From the first equation, we get $\frac{2}{u} = \frac{1}{v} + \frac{1}{v+1}$, hence $u = \frac{2v(v+1)}{2v+1}$. From the second equation, we get $\frac{1}{v} + \frac{1}{u+v} = 1 + \frac{1}{v+1}$, substituting $u$, and rearranging it as an equation about $v$,
$$4v^3 + 5v^2 - 4v - 4 = 0.$$
Let $f(t) = 4t^3 + 5t^2 - 4t - 4$, then $f'(t) = 12t^2 + 10t - 4$, which is negative at first and positive later on $[0, +\infty)$, so $f(t)$ first decreases and then increases on $[0, +\infty)$. Since $f(0) = -4 < 0$, $f(1) = 1 > 0$, therefore, $f$ has a unique solution $v \in (0, 1)$ on $[0, +\infty)$, and $u$ is also uniquely determined.

(5) Let $u, v > 0$ be the solution satisfying the system of equations in (4), and let $\frac{2}{u} + \frac{1}{u+v} = k$.

Using the inequality of arithmetic and geometric means, we have:
$$\begin{align*} F(x_1, x_6, x_8) &= \left(\frac{x_1}{u}\right)^4 \left(\frac{x_6}{v}\right)^2 x_8 \left(\frac{x_1+x_6}{u+v}\right)^2 \left(\frac{x_6+x_8}{v+1}\right)^2 u^4 v^2 (u+v)^2 (v+1)^2 \\ &\le \frac{u^4 v^2 (u+v)^2 (v+1)^2}{11^{11}} \left(\frac{4}{u} x_1 + \frac{2}{v} x_6 + x_8 + \frac{2}{u+v} (x_1+x_6) + \frac{2}{v+1} (x_6+x_8)\right)^{11} \\ &= \frac{u^4 v^2 (u+v)^2 (v+1)^2}{11^{11}} \left(\left(\frac{4}{u} + \frac{2}{u+v}\right) x_1 + \left(\frac{2}{v} + \frac{2}{u+v} + \frac{2}{v+1}\right) x_6 + \left(1 + \frac{2}{v+1}\right) x_8\right)^{11} \\ &= \frac{u^4 v^2 (u+v)^2 (v+1)^2}{11^{11}} (2kx_1 + 2kx_6 + kx_8)^{11} = \frac{u^4 v^2 (u+v)^2 (v+1)^2 k^{11}}{11^{11}}. \end{align*}$$
The equality above holds if and only if $x_1 : x_6 : x_8 = u : v : 1$, that is, $x_8 = \frac{1}{2u+2v+1}$,
$x_6 = \frac{v}{2u+2v+1}$, $x_1 = \frac{u}{2u+2v+1}$. Therefore, $F$ attains its maximum value if and only if $x_2 = x_3 =$
$x_5 = x_7 = x_9 = x_{11} = 0$, $x_1 = x_4 = \frac{u}{2u+2v+1}$, $x_6 = x_{10} = \frac{v}{2u+2v+1}$, $x_8 = \frac{1}{2u+2v+1}$. At this point,
$x_6 = vx_8 < x_8$. $\square$