Giải thích các bước giải:
a.$A=\sqrt{x-2018}+\sqrt{2019-x}\ge \sqrt{x-2018+2019-x}=1$
Dấu = xảy ra khi $x=2019$
b.Ta có:
$\sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}$
$=\sqrt{x(x+y+z)+yz}+\sqrt{y(x+y+z)+zx}+\sqrt{z(x+y+z)+xy}$
$=\sqrt{x^2+xy+xz+yz}+\sqrt{y^2+yx+yz+zx}+\sqrt{z^2+xz+zy+xy}$
$=\sqrt{(x+y)(x+z)}+\sqrt{(y+z)(y+x)}+\sqrt{(z+x)(z+y)}$
$\ge \sqrt{(x+\sqrt{yz})^2}+\sqrt{(y+\sqrt{xz})^2}+\sqrt{(z+\sqrt{xy})^2}$
$\ge x+\sqrt{yz}+y+\sqrt{xz}+z+\sqrt{xy}$
$\ge x+y+z+\sqrt{xy}+\sqrt{yz}+\sqrt{xz}$
$\ge 1+\sqrt{xy}+\sqrt{yz}+\sqrt{xz}$