Question

Chứng minh:

a. x² + xy + y² + 1 > 0 với mọi x, y

b. x² + 4y² + z² - 2x - 6z + 8y + 15 > 0 Với mọi x, y, z

Answer 1

⇒(x−1)^2+4(y+1)^2+(z−3)^2≥0

x^2+4y^2+z^2-2x-6z+8y+15

=x^2+4y^2+z^2-2x-6z+8y+1+1+4+9

=(x^2-2x+1)+(4y^2+8y+4)+(z^2-6z+9)+1

=(x-1)^2+4(y+1)^2+(z-3^)2+1

Ta thấy:(x−1)^2≥0

              4(y+1)^2≥0

             (z−3)^ 2≥0

{(x−1)^24(y+1)^2(z−3)^2≥0

⇒(x−1)^2+4(y+1)^2+(z−3)^2≥0

⇒(x−1)2+4(y+1)2+(z−3)2+1≥0+1=1>0

Answer 2

\(x^2+xy+y^2+1.=x^2+2.x.\dfrac{y}{2}+\left(\dfrac{y}{2}\right)^2+\dfrac{3}{4}y^2+1.\\ =\left(x+\dfrac{y}{2}\right)^2+\dfrac{3}{4}y^2+1>0\forall x;y\in R.\\ \Rightarrow x^2+xy+y^2+10\forall x;y\in R.\)

Answer 3

Kkk