Ta có P = \(\frac{x}{3+x}+\frac{y}{3+y}+\frac{z}{3+z}\)
= \(1-\frac{3}{x+3}+1-\frac{3}{y+3}+1-\frac{3}{z+3}\)
= \(3-\left(\frac{3}{x+3}+\frac{3}{y+3}+\frac{3}{z+3}\right)\)
\(\le3-\frac{\left(3\sqrt{3}\right)^2}{x+y+z+9}=3-\frac{27}{12}\)
= \(\frac{3}{4}\)
Đạt được khi x = y = z = 1