2022 CGMO

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 $a_3^{\prime }=0$, $a_4^{\prime }=a_3+a_4$, and let $a_i^{\prime }=a_i$ for other $i$. This maintains the sum. We prove that $F(a_1,a_2,\cdots ,a_{11})<F(a_1^{\prime },a_2^{\prime },\cdots ,a_{11}^{\prime })$, which is equivalent to $$(a_2+a_3+a_4)(a_3+a_4)(a_4+a_5+a_6)<(a_2^{\prime }+a_3^{\prime }+a_4^{\prime })(a_3^{\prime }+a_4^{\prime })(a_4^{\prime }+a_5^{\prime }+a_6^{\prime }),$$ and since $a_3+a_4=a_3^{\prime }+a_4^{\prime }$ and $a_4<a_4^{\prime }$, 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 $a_1^{\prime }=a_1+a_2$, $a_2^{\prime }=0$, and let $a_i^{\prime }=a_i$ for other $i$. This maintains the sum. We prove that $F(a_1,a_2,a_4,\cdots ,a_{10})<F(a_1^{\prime },a_2^{\prime },a_4^{\prime },\cdots ,a_{10}^{\prime })$. $$\iff a_1(a_{10}+a_1)(a_1+a_2)(a_2+a_4)<a_1^{\prime }(a_{10}^{\prime }+a_1^{\prime })(a_1^{\prime }+a_2^{\prime })(a_2^{\prime }+a_4^{\prime }).$$ Since $a_1<a_1^{\prime }$ and $a_{10}+a_1<a_{10}^{\prime }+a_1^{\prime }$, we only need to prove: $$a_1(a_1+a_2)(a_2+a_4)\le a_1^{\prime }(a_1^{\prime }+a_2^{\prime })(a_2^{\prime }+a_4^{\prime })=(a_1+a_2)(a_1+a_2)a_4,$$ $$\iff a_1(a_2+a_4)\le a_4(a_1+a_2),\iff a_1{a}_2\le a_4{a}_2,$$ which holds. If $a_1\ge a_4$, let $a_4^{\prime }=a_4+a_2$, $a_2^{\prime }=0$, and for other $i$, let $a_i^{\prime }=a_i$. We also have: $$F(a_1,a_2,a_4,\cdots ,a_{10})<F(a_1^{\prime },a_2^{\prime },a_4^{\prime },\cdots ,a_{10}^{\prime }).$$ 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 $a_1^{\prime }=a_4^{\prime }=\frac{1}{2}(a_1+a_4)$, $a_6^{\prime }=a_{10}^{\prime }=\frac{1}{2}(a_6+a_{10})$, $a_8^{\prime }=a_8$, keeping the sum unchanged. By the inequality of arithmetic and geometric means, a_1^2 a_4^2 \le a'_1^2 a'_4^2, \quad (a_4 + a_6)(a_{10} + a_1) \le (a'_4 + a'_6)(a'_{10} + a'_1), $$a_6{a}_{10}\le a_6^{\prime }{a}_{10}^{\prime },(a_6+a_8)(a_8+a_{10})\le (a_6^{\prime }+a_8^{\prime })(a_8^{\prime }+a_{10}^{\prime }).$$ Also, $a_8=a_8^{\prime }$. Multiplying these equations gives: $$F(a_1,a_4,\dots ,a_{10})\le F(a_1^{\prime },a_4^{\prime },\dots ,a_{10}^{\prime }).$$ 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)): $$\frac{2}{u}+\frac{1}{u+v}=\frac{1}{v}+\frac{1}{u+v}+\frac{1}{v+1}=1+\frac{2}{v+1},u,v>0.$$ 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 $f^{\prime }(t)=12t^2+10t-4$, 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: $$\begin{array}{rl}F(x_1,x_6,x_8)& ={\left(\frac{x_1}{u}\right)}^4{\left(\frac{x_6}{v}\right)}^2{x}_8{\left(\frac{x_1+x_6}{u+v}\right)}^2{\left(\frac{x_6+x_8}{v+1}\right)}^2{u}^4{v}^2(u+v{)}^2(v+1{)}^2\\ & \le \frac{u^4{v}^2(u+v{)}^2(v+1{)}^2}{11^{11}}{\left(\frac{4}{u}{x}_1+\frac{2}{v}{x}_6+x_8+\frac{2}{u+v}(x_1+x_6)+\frac{2}{v+1}(x_6+x_8)\right)}^{11}\\ & =\frac{u^4{v}^2(u+v{)}^2(v+1{)}^2}{11^{11}}{\left(\left(\frac{4}{u}+\frac{2}{u+v}\right)x_1+\left(\frac{2}{v}+\frac{2}{u+v}+\frac{2}{v+1}\right)x_6+\left(1+\frac{2}{v+1}\right)x_8\right)}^{11}\\ & =\frac{u^4{v}^2(u+v{)}^2(v+1{)}^2}{11^{11}}(2k{x}_1+2k{x}_6+k{x}_8{)}^{11}=\frac{u^4{v}^2(u+v{)}^2(v+1{)}^2{k}^{11}}{11^{11}}.\end{array}$$ 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, $x_6=v{x}_8<x_8$. $\square$