1/ Có (x+y)*6=(x+y)*(\frac{2}{x}+\frac{3}{y}) \geq (\sqrt{2}+\sqrt{3})2 (bđt B.C.S)\Rightarrow S \geq \frac{(\sqrt{2}+\sqrt{3})2}{6}Dấu = có khi x=\frac{2+\sqrt{6}}{6}; y=\frac{3+\sqrt{6}}{6}
1/ Có
$(x+y)*6=(x+y)*(\frac{2}{x}+\frac{3}{y}) \geq (\sqrt{2}+\sqrt{3})
^2
$ (bđt B.C.S)
$\Rightarrow S \geq \frac{(\sqrt{2}+\sqrt{3})
^2}{6}
$Dấu
$=
$ có khi
$x=\frac{2+\sqrt{6}}{6}; y=\frac{3+\sqrt{6}}{6}
$