Question

cm vs mọi x,y ta có: 4x² +4xy + 4y²  >6y -4

Answer 1

\(4x^2+4xy+4y^2-6y+4=\left(4x^2+4xy+y^2\right)+3\left(y^2-2y+1\right)+1\)

\(=\left(2x+y\right)^2+3\left(y-1\right)^2+1>0\)

Do đó \(4x^2+4xy+4y^2>6y+4\)

Answer 2

Ta có:

4x² + 4xy + 4y² - 6y + 4

= 4x² + 4xy + y² + 3y² + 6y + 4

= (4x² + 4xy + y²) + (3y² + 6y + 4)

= (2x + y)² + 3(y² + 2y + 4/3)

= (2x + y)² + 3(y² + 2y + 1 + 1/3)

= (2x + y)² + 3(y + 1)² + 1

Do (2x + y)² >= 0

(y + 1)² >= 0

Suy ra (2x + y)² + 3(y + 1)² >= 0

Suy ra (2x + y)² + 3(y + 1)² + 1 > 0

Suy ra 4x² + 4xy + 4y² - 6y + 4 > 0

Hay 4x² + 4xy + 4y² > 6y - 4