Áp dụng bất đẳng thức Bunhiacopxki :
\(T^2=\left|x+y\right|^2\le\left|\left(x^2+y^2\right)\left(1+1\right)\right|=\left|4\cdot2\right|=8\)
Do đó \(T^2\le8\)
\(\Leftrightarrow-\sqrt{8}\le T\le\sqrt{8}\)
\(\Leftrightarrow-2\sqrt{2}\le T\le2\sqrt{2}\)
+) Vậy \(T_{min}=-2\sqrt{2}\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x+y=-2\sqrt{2}\\x=y\end{matrix}\right.\)\(\Leftrightarrow x=y=-\sqrt{2}\)
+) Vậy \(T_{max}=2\sqrt{2}\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x+y=2\sqrt{2}\\x=y\end{matrix}\right.\)\(\Leftrightarrow x=y=\sqrt{2}\)