\(I^2=\left(\frac{3}{\sqrt{2}}.\sqrt{2}x+\sqrt{2}.y\right)^2\le\left(\frac{9}{2}+2\right)\left(2x^2+y^2\right)=\frac{169}{4}\)
\(\Rightarrow I\le\sqrt{\frac{169}{4}}=\frac{13}{2}\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}2x^2+y^2=\frac{13}{2}\\\frac{2x}{3}=\frac{y}{\sqrt{2}}\end{matrix}\right.\)