a/ Áp dụng BĐT Cô si cho 2 số dương:
$2x+3y≥2\sqrt{2x.3y}=2\sqrt{6xy}$
$\dfrac{2}{x}+\dfrac{3}{y}≥2\sqrt{\dfrac{2}{x}.\dfrac{3}{y}}=2\sqrt{\dfrac{6}{xy}}$
$3x+\dfrac{3}{x}≥2\sqrt{3x.\dfrac{3}{x}}=6$
$→(2x+3y)(\dfrac{2}{x}+\dfrac{3}{y})(3x+\dfrac{3}{x})≥2\sqrt{6xy}.2\sqrt{\dfrac{6}{xy}}.6=144$
$→$ Dấu "=" xảy ra khi $\begin{cases}2x=3y\\\dfrac{2}{x}=\dfrac{3}{y}\\3x=\dfrac{3}{x}\end{cases}$
$↔\begin{cases}2x=3y\\\dfrac{2}{x}=\dfrac{3}{y}\\x=1(\text{vì $x>0$})\end{cases}$
$↔\begin{cases}x=1\\y=\dfrac{3}{2}\end{cases}$ (tm)
