Question

CM: x² + xy + y² + 1  > 0  với mọi x , y

Answer 1

x² + xy + y²  + 1 = (x² + 2.x. \(\frac{y}{2}\) + (\(\frac{y}{2}\))² ) + \(\frac{3y^2}{4}\) + 1 = (x + \(\frac{y}{2}\))² + \(\frac{3y^2}{4}\) + 1 \(\ge\) 0 + 0 + 1 = 1> 0 với mọi x; y

Answer 2

Ta có:

x²+xy+y²+1=x²+xy+1/4.y²+3/4.y²+1=(x+1/2.y)²+3/4.y²+1

Mà (x+1/2.y)² \(\ge\)0

3/4.y²>=0

1>0

Suy ra (x+1/2.y)²+3/4.y²+1>0

Hay x²+xy+y²+1>0(đpcm)

Answer 3

≥0

3/4.y²>=0

1>0

Suy ra (x+1/2.y)²+3/4.y²+1>0

Hay x²+xy+y²+1>0(đpcm)