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

By AM-GM inequality
$\frac{3^{y_{1}}+3^{y_{2}}+3^{y_{3}}}{3} \geq\left(3^{y_{1}} \cdot 3^{y_{2}} \cdot 3^{y_{3}}\right)^{\frac{1}{3}}$
$\Rightarrow 3^{y_{1}}+3^{y_{2}}+3^{y_{3}} \geq 3\left(\frac{{y_{1}}+y_{2}+y_{3}}{3}\right)$
$\Rightarrow 3^{y_{1}}+3^{y_{2}}+3^{y_{3}} \geq 3^{4}$
$\Rightarrow \log 3\left(3^{y_{1}}+3^{y_{2}}+3^{y_{3}}\right) \geq 4$
$\Rightarrow m=4$ ,
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 $)$
$x_{1}+x_{2}+x_{3} \geq 3\left(x_{1} x_{2} x_{3}\right)^{\frac{1}{3}}$
$3 \geq\left(x_{1} x_{2} x_{3}\right)^{\frac{1}{3}}$
$\therefore \log _{3} 3 \geq \frac{1}{3} \log \left(x_{1} x_{2} x_{3}\right)$
$\therefore 3 \geq P$
$\Rightarrow M=3$
$\therefore \log _{2} m^{3}+\log _{3} m^{2}=\log _{2} 4^{3}+\log _{3} 3^{2}$
$=6+2=8$