Đáp án:
Giải thích các bước giải:
$P \geq \dfrac{1}{x^2+y^2+z^2}+\dfrac{3}{4}.\dfrac{9}{xy+yz+zx}$
$P \geq \dfrac{1}{x^2+y^2+z^2}+\dfrac{27}{4}.\dfrac{1}{xy+yz+zx}$
$P \geq \dfrac{1}{x^2+y^2+z^2}+\dfrac{1}{xy+yz+zx}+\dfrac{1}{xy+yz+zx}+\dfrac{19}{4}.\dfrac{1}{xy+yz+zx}$
$P \geq \dfrac{9}{x^2+y^2+z^2+xy+yz+zx+xy+yz+zx}+\dfrac{57}{4}.\dfrac{1}{3(xy+yz+zx)}$
$P \geq \dfrac{9}{(x+y+z)^2}+\dfrac{57}{4}.\dfrac{1}{(x+y+z)^2}=\dfrac{93}{4.2019^2}$
Dấu "=" xảy ra khi $x=y=z=\dfrac{2019}{3}$
