Question

Let $m$ be the minimum possible value of ${\log }_3\left(3^{y_1}+3^{y_2}+3^{y_3}\right)$, 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 $\left({\log }_3{x}_1+{\log }_3{x}_2+{\log }_3{x}_3\right)$, where $x_1,x_2,x_3$ are positive real numbers for which $x_1+x_2+x_3=9$. Then the value of ${\log }_2\left(m^3\right)+{\log }_3\left(M^2\right)$ is_______.

Answer 1

Answer: 8

By AM-GM inequality
$\frac{3^{y_1}+3^{y_2}+3^{y_3}}{3}\ge {\left(3^{y_1}\cdot 3^{y_2}\cdot 3^{y_3}\right)}^{\frac{1}{3}}$
$\Rightarrow 3^{y_1}+3^{y_2}+3^{y_3}\ge 3\left(\frac{y_1+y_2+y_3}{3}\right)$
$\Rightarrow 3^{y_1}+3^{y_2}+3^{y_3}\ge 3^4$
$\Rightarrow \log 3\left(3^{y_1}+3^{y_2}+3^{y_3}\right)\ge 4$
$\Rightarrow m=4$ ,
Similarly
For ${\log }_3{x}_1+{\log }_3{x}_2+{\log }_3{x}_3={\log }_3\left(x_1{x}_2{x}_3\right)=P($ say $)$
$x_1+x_2+x_3\ge 3{\left(x_1{x}_2{x}_3\right)}^{\frac{1}{3}}$
$3\ge {\left(x_1{x}_2{x}_3\right)}^{\frac{1}{3}}$
$\therefore {\log }_{3}3\ge \frac{1}{3}\log \left(x_1{x}_2{x}_3\right)$
$\therefore 3\ge P$
$\Rightarrow M=3$
$\therefore {\log }_2{m}^3+{\log }_3{m}^2={\log }_2{4}^3+{\log }_3{3}^2$
$=6+2=8$