a) Có x - 2y = 1 => x = 1 + 2y
Thay vào ta có:
B = \(\left(1+2y\right)^2-2y^2+2020\)
= 1 + 4y + \(4y^2\) - \(2y^2\) + 2020
= 4y +\(2y^2\) + 2021
= 2.\(\left(y^2+2y+1\right)\) + 2019
= \(2\left(y+1\right)^2+2019\) \(\ge2019\)
Dấu "=" xảy ra <=> y = -1; x = -1
b) Có x+y+z = 3 => \(\left(x+y+z\right)^2=9\)
=> \(x^2+y^2+z^2+2\left(xy+yz+zx\right)=9\)
Ta có: \(\left(x+y\right)^2\ge0=>2xy\le x^2+y^2\)
Tương tự: \(2yz\le y^2+z^2;2zx\le z^2+x^2\)
=> 2(xy + yz + zx) \(\le\) \(2x^2+2y^2+2z^2\)
=> xy + yz + zx \(\le\) \(x^2+y^2+z^2\)
=> 3(xy + yz + zx) \(\le\)\(\left(x+y+z\right)^2\) = 9
=> 2(xy + yz + zx) \(\le\) 6
=> \(x^2+y^2+z^2\ge3\)
Dấu "=" xảy ra <=> x = y = z = 1
