Giải thích các bước giải:
$A=\dfrac{1}{2a}+\dfrac{3}{b}$
$\rightarrow A=\dfrac{1}{2a}+\dfrac{3^2}{3b}$
$\rightarrow A(2a+3b)=(\dfrac{1}{2a}+\dfrac{3^2}{3b})(2a+3b)$
$\rightarrow A(2a+3b)\ge (\sqrt{\dfrac{1}{2a}.2a}+\sqrt{\dfrac{3^2}{3b}.3b})^2$
$\rightarrow A(2a+3b)\ge (1+3)^2$
$\rightarrow A\ge \dfrac{16}{2a+3b} \ge \dfrac{12}{2a+3b}$