\(P=\frac{x}{x+3}+\frac{y}{y+3}+\frac{z}{z+3}=1-\frac{3}{x+3}+1-\frac{3}{y+3}+1-\frac{3}{z+3}\)
\(P=3-3\left(\frac{1}{x+3}+\frac{1}{y+3}+\frac{1}{z+3}\right)\le3-3.\frac{9}{x+y+z+9}=3-\frac{27}{12}=\frac{3}{4}\)
\(\Rightarrow P_{max}=\frac{3}{4}\) khi \(x=y=z=1\)