Giải bài 2:
Ta có:
\(x^2+y^2\ge\dfrac{\left(x+y\right)^2}{2}\)
\(\Leftrightarrow x^2+y^2-\left(x+y\right)\ge\dfrac{\left(x+y\right)^2}{2}-\left(x+y\right)\\ \Leftrightarrow x\left(x-1\right)+y\left(y-1\right)\ge\dfrac{6^2}{2}-6\\ \Leftrightarrow x\left(x-1\right)+y\left(y-1\right)\ge12\left(đpcm\right)\)
Dấu "=" xảy ra khi x = y = 3.



