Do \(x^2+y^2\ge0\) \(\forall x;y\Rightarrow x+y\ge0\)
Lại có \(x^2+y^2\ge\frac{\left(x+y\right)^2}{2}\Rightarrow x+y\ge\frac{\left(x+y\right)^2}{2}\)
\(\Rightarrow2\left(x+y\right)-\left(x+y\right)^2\ge0\Rightarrow\left(x+y\right)\left(2-\left(x+y\right)\right)\ge0\)
- Nếu \(x+y=0\Rightarrow x+y< 2\) BĐT đúng
- Nếu \(x+y>0\Rightarrow2-\left(x+y\right)\ge0\Rightarrow x+y\le2\)
Vậy \(x+y\le2\)