The 16th Japanese Mathematical Olympiad - The Final Round

First we prove that, for any positive real numbers $p_1,\dots ,p_n,q_1,\dots ,q_n,r_1,\dots ,r_n$, the following inequality holds: $$(p_1^3+\cdots +p_n^3)(q_1^3+\cdots +q_n^3)(r_1^3+\cdots +r_n^3)\ge (p_1{q}_1{r}_1+\cdots +p_n{q}_n{r}_n{)}^3.$$ Indeed, by using Cauchy-Schwarz inequality repeatedly, we obtain $$\begin{array}{rl}& (p_1^3+\cdots +p_n^3)(q_1^3+\cdots +q_n^3)(r_1^3+\cdots +r_n^3)(p_1{q}_1{r}_1+\cdots +p_n{q}_n{r}_n)\\ \ge & ((p_1^{\frac{3}{2}}{q}_1^{\frac{3}{2}}+\cdots +p_n^{\frac{3}{2}}{q}_n^{\frac{3}{2}})(p_1^{\frac{1}{2}}{q}_1^{\frac{1}{2}}{r}_1^2+\cdots +p_n^{\frac{1}{2}}{q}_n^{\frac{1}{2}}{r}_n^2){)}^2\\ \ge & ((p_1{q}_1{r}_1+\cdots +p_n{q}_n{r}_n{)}^2{)}^2\\ =& (p_1{q}_1{r}_1+\cdots +p_n{q}_n{r}_n{)}^4\end{array}$$ and hence the desired inequality.

Let $t=6^{-\frac{1}{3}}$. Using this inequality and AM-GM inequality, $$\begin{array}{rl}& (x_1^3+x_2^3+x_3^3+1)(y_1^3+y_2^3+y_3^3+1)(z_1^3+z_2^3+z_3^3+1)\\ & =(x_1^3+x_2^3+x_3^3+t^3+t^3+t^3+t^3+t^3+t^3)\\ & \times (t^3+t^3+t^3+y_1^3+y_2^3+y_3^3+t^3+t^3+t^3)\\ & \times (t^3+t^3+t^3+t^3+t^3+t^3+z_1^3+z_2^3+z_3^3)\\ & \ge (t^2{x}_1+t^2{x}_2+t^2{x}_3+t^2{y}_1+t^2{y}_2+t^2{y}_3+t^2{z}_1+t^2{z}_2+t^2{z}_3{)}^3\\ & =\frac{1}{36}{\left((x_1+y_1+z_1)+(x_2+y_2+z_2)+(x_3+y_3+z_3)\right)}^3\\ & \ge \frac{3^3}{36}(x_1+y_1+z_1)(x_2+y_2+z_2)(x_3+y_3+z_3).\end{array}$$ Equality holds if and only if equality holds for each Cauchy-Schwarz and AM-GM, namely $x_1=x_2=x_3=y_1=y_2=y_3=z_1=z_2=z_3=t$.

Therefore, the maximum value of $A$ is $\frac{3}{4}$ and equality holds if $x_1=x_2=x_3=y_1=y_2=y_3=z_1=z_2=z_3=t$.