Let $m$ be the minimum possible value of $\log_3(3^{y_1} + 3^{y_2} + 3^{y_3})$, where $y_1, y_2, y_3$ are real numbers for which $y_1 + y_2 + y_3 = 9$. Let $M$ be the maximum possible value of $(\log_3 x_1 + \log_3 x_2 + \log_3 x_3)$, where $x_1, x_2, x_3$ are positive real numbers for which $x_1 + x_2 + x_3 = 9$. Then the value of $\log_2(m^3) + \log_3(M^2)$ is ______

Let m be the minimum possible value of log₃(3^(y₁) + 3^(y₂) + 3^(y₃)), where y₁, y₂, y₃ are real numbers for which y₁ + y₂ + y₃ = 9. Let M be the maximum possible value of (log₃ x₁ + log₃ x₂ + log₃ x₃), where x₁, x₂, x₃ are positive real numbers for which x₁ + x₂ + x₃ = 9. Then the value of log₂(m³) + log₃(M²) is ______

$\text{Given: } y_1 + y_2 + y_3 = 9,\quad y_1, y_2, y_3 \in \mathbb{R}$

$x_1 + x_2 + x_3 = 9,\quad x_1, x_2, x_3 > 0$

$\text{To find: } \log_2(m^3) + \log_3(M^2)$

$m = \min \log_3\!\left(3^{y_1} + 3^{y_2} + 3^{y_3}\right)$

$M = \max\left(\log_3 x_1 + \log_3 x_2 + \log_3 x_3\right)$

$\text{AM-GM: } \dfrac{a_1+a_2+a_3}{3} \ge (a_1 a_2 a_3)^{1/3}, \quad a_i > 0$

$\text{Equality } \iff a_1 = a_2 = a_3$

$\dfrac{3^{y_1}+3^{y_2}+3^{y_3}}{3} \ge (3^{y_1}\cdot 3^{y_2}\cdot 3^{y_3})^{1/3}$

$= \left(3^{\,y_1+y_2+y_3}\right)^{1/3}$

$3^{9/3} = 3^3 = 27 \ \Rightarrow\ 3^{y_1}+3^{y_2}+3^{y_3} \ge 3\cdot 27 = 81$

$\text{Equality: } y_1=y_2=y_3=3 \ \Rightarrow\ m = \log_3 81 = 4$

$\log_3 x_1 + \log_3 x_2 + \log_3 x_3 = \log_3(x_1 x_2 x_3)$

$\dfrac{x_1+x_2+x_3}{3} \ge (x_1 x_2 x_3)^{1/3}$

$3 \ge (x_1 x_2 x_3)^{1/3} \ \Rightarrow\ x_1 x_2 x_3 \le 27$

$M = \log_3(x_1 x_2 x_3)_{\max} = \log_3 27 = 3$

$\log_2(m^3) = \log_2(4^3) = \log_2 64 = \log_2 2^6 = 6$

$\log_3(M^2) = \log_3(3^2) = \log_3 9 = 2$

$\log_2(m^3) + \log_3(M^2) = 6 + 2 = 8$