QJEEAdv20Maae94: JEE Advanced 2020 Mathematics - Class 11 JEEAdv20Maae94 Solution: Let m be the minimum possible value of , where are real numbers …

For real numbers y₁, y₂, y₃, the quantities ${{3^{{y_1}}}}$ , ${{3^{{y_2}}}}$ and ${{3^{{y_3}}}}$ are positive real numbers, so according to the AM-GM inequality, we have

${{{3^{{y_1}}} + {3^{{y_2}}} + {3^{{y_3}}}} \over 3} \ge {({3^{{y_1}}}\,.\,{3^{{y_2}}}\,.\,{3^{{y_3}}})^{{1 \over 3}}}$

$\Rightarrow {3^{{y_1}}} + {3^{{y_2}}} + {3^{{y_3}}} \ge 3{({3^{{y_1}}}\,.\,{3^{{y_2}}}\,.\,{3^{{y_3}}})^{{1 \over 3}}}$

On applying logarithm with base '3', we get

${\log _3}({3^{{y_1}}} + {3^{{y_2}}} + {3^{{y_3}}}) \ge \left[ {1 + {1 \over 3}({y_1} + {y_2} + {y_3}} \right.)]$

= 1 + 3

= 4

{ $\because$ ${{y_1} + {y_2} + {y_3}}$ = 9}

$\therefore$ m = 4

Now, for positive real numbers x₁, x₂ and x₃ according to AM-GM inequality, we have

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

On applying logarithm with base '3', we get

${\log _3}\left( {{{{x_1} + {x_2} + {x_3}} \over 3}} \right) \ge {1 \over 3}$ $({\log _3}{x_1} + {\log _3}{x_2} + {\log _3}{x_3})$

$\Rightarrow$ $1 \ge {1 \over 3}\left( {{{\log }_3}{x_1} + {{\log }_3}{x_2} + {{\log }_3}{x_3}} \right)$

{ $\because$ x₁ + x₂ + x₃ = 9}

$\therefore$ M = 3

Now, ${\log _2}({m^3}) + {\log _3}({M^2})$

$= 3lo{g_2}(4) + 2lo{g_2}(3)$ = 6 + 2 = 8