Ta có:
\(P=\left(\frac{1}{x^2+y^2+z^2}+\frac{1}{3xy}+\frac{1}{3yz}+\frac{1}{3zx}\right)+\frac{5}{12}\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)\)
\(\ge\frac{\left(1+1+1+1\right)^2}{x^2+y^2+z^2+3xy+3yz+3zx}+\frac{5}{12}.\frac{\left(1+1+1\right)^2}{xy+yz+zx}\)
\(=\frac{16}{\left(x+y+z\right)^2+\left(xy+yz+zx\right)}+\frac{5}{12}.\frac{9}{xy+yz+zx}\)
\(\ge\frac{16}{\left(x+y+z\right)^2+\frac{\left(x+y+z\right)^2}{3}}+\frac{5}{12}.\frac{9}{\frac{\left(x+y+z\right)^2}{3}}\)
\(=\frac{93}{4\left(x+y+z\right)^2}=\frac{93}{4\left(2019\right)^2}\)
Dấu "=" xảy ra <=> x = y = z = 2019/3.
