\(2x+3y=\dfrac{1}{\sqrt{2}}.2.\sqrt{2}x+\dfrac{1}{\sqrt{3}}.3\sqrt{3}y=7\le\sqrt{[\left(\dfrac{2}{\sqrt{2}}\right)^2+\left(\dfrac{3}{\sqrt{3}}\right)^2]\left(2x^2+3y^2\right)}=\sqrt{5\left(2x^2+3y^2\right)}\Leftrightarrow2x^2+3y^2\ge\dfrac{49}{5}\)
\(dấu"="\Leftrightarrow x=y=\dfrac{7}{5}\)