B = x2 + y2 + 2x + 2y + 3
B = (x2 + 2x + 1) + (y2 + 2y + 1) + 1
B = (x + 1)2 + (y + 1)2 + 1
Vì (x + 1)2 \(\ge\) 0 \(\forall\) x; (y + 1)2 \(\ge\) 0 \(\forall\) y
=> (x + 1)2 + (y + 1)2 \(\ge\) 0 \(\forall\) x, y
=> B \(\ge\) 1 \(\forall\) x, y (đpcm)
Bạn bổ sung ở cuối là B\(\ge\) 1 \(\forall\) x, y thì sẽ > 0 \(\forall\) x, y nhé