Câu 48 này giải như thế nào vậy mọi người ơi?!@_@

$\begin{array}{l}
2x + y{.4^{x + y - 1}} \ge 3\\
 \Leftrightarrow y{.4^{x + y - 1}} \ge 3 - 2x\\
 \Leftrightarrow y{.4^{x - 1}}{.4^y} \ge 3 - 2x\\
 \Leftrightarrow y{.4^y} \ge \left( {3 - 2x} \right){.4^{1 - x}}\\
 \Leftrightarrow y{.2^{2y}} \ge \left( {3 - 2x} \right){.2^{2 - 2x}}\\
 \Leftrightarrow 2y{.2^{2y}} \ge \left( {3 - 2x} \right){.3^{3 - 2x}}\left( * \right)\\
 \bullet T{H_1}:3 - 2x \le 0 \Rightarrow x \ge \dfrac{3}{2} \Rightarrow VT > 0,VP < 0\\
 \Rightarrow \text{Đúng} \forall y \ge 0\\
P = {x^2} + {y^2} + 6x + 4y \ge {\left( {\dfrac{3}{2}} \right)^2} + {0^2} + 6.\dfrac{3}{2} + 4.0 = \dfrac{{45}}{4}\\
 \Rightarrow {\min _1}P = \dfrac{{45}}{4}\\
 \bullet T{H_2}:3 - 2x > 0 \Leftrightarrow x < \dfrac{3}{2} \to \left( {3 - 2x} \right){.2^{3 - 2x}} > 0\\
f\left( t \right) = t{.2^t}\forall t > 0 \Rightarrow f'\left( t \right) = {2^t} + t{2^t}\ln 2 > 0\\
 \Rightarrow \text{Hàm số đông biến}\\
 \Rightarrow f\left( {2y} \right) \ge f\left( {3 - 2x} \right)\\
 \Leftrightarrow 2y \ge 3 - 2x\\
 \Leftrightarrow y \ge \dfrac{3}{2} - x\\
 \Rightarrow P = {x^2} + {y^2} + 6x + 4y \ge {x^2} + {\left( {\dfrac{3}{2} - x} \right)^2} + 6x + 4\left( {\dfrac{3}{2} - x} \right)\\
 = 2{x^2} - 3x + \dfrac{9}{4} + 6x + 6 - 4x\\
 = 2{x^2} + x + \dfrac{{33}}{4} \ge 2{\left( {x + \dfrac{1}{4}} \right)^2} + \dfrac{{65}}{8} \ge \dfrac{{65}}{8}\\
 \Rightarrow {\min _2}P = \dfrac{{65}}{8}\\
 \Rightarrow \min P = \dfrac{{65}}{8} \to C
\end{array}$