[tex](x^2-y^2)(x^3-y^3)\geq 0=>x^5+y^5\geq x^2y^2(x+y)=>x^5+y^5+xy\geq xy(1+xy(x+y))[/tex]
=>[tex]\frac{xy}{x^5+y^5+xy}\leq \frac{1}{xy(x+y)+1}=\frac{1}{xy(x+y)+xuz}=\frac{1}{xy(x+y+z)}[/tex]
tượng tự .................................
=> P[tex]\leq \frac{1}{xy(x+y+z)}+ \frac{1}{xz(x+y+z)}+ \frac{1}{yz(x+y+z)}=\frac{x+y+z}{xyz(x+z+y)}=1[/tex]
Vậy gtln của P=1
khi x=y=z