Question

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 ${\log }_2\left(m^3\right)+{\log }_3\left(M^2\right)$ is

Answer 1

using $A.M.\ge G.M.$
$3^{y_1}+3^{y_2}+3^{y_3}\ge 3\cdot \left(3^{y_1+y_2+y_3}{)}^{1/3}\right[\because y_1+y_2+y_3=9]$
$\therefore 3^{y_1}+3^{y_2}+3^{y_3}\ge 3\cdot (3^9{)}^{1/3}$
$3^{y_1}+3^{y_2}+3^{y_3}\ge 3^4$
$\Rightarrow m={\log }_3{3}^4=4$

Again, using $A.M.\ge G.M.$
$\frac{x_1+x_2+x_3}{3}\ge (x_1\cdot x_2\cdot x_3{)}^{1/3}$
$=\frac{9}{3}\ge (x_1\cdot x_2\cdot x_3{)}^{1/3}$
$\Rightarrow 27\ge (x_1\cdot x_2\cdot x_3)$
$M={\log }_3{x}_1+{\log }_3{x}_2+{\log }_3{x}_3$
$M={\log }_3\left(x_1{x}_3{x}_3\right)={\log }_{3}27=3$
$\therefore {\log }_2(m{)}^3+{\log }_3(M{)}^2={\log }_2\left(2^6\right)+{\log }_3\left(3^2\right)=6+2=8$