Toán 8 Tìm Max(P)

Có
- [tex]\frac{x}{2x^2+y^2+5}=\frac{x}{(x^2+y^2)+(x^2+1)+4} \leq \frac{x}{2xy+2x+4} \\ \frac{2y}{6y^2+z^2+6}=\frac{2y}{(4y^2+z^2)+(2y^2+2)+4} \leq -\frac{2y}{4yz+4y+4}=\frac{y}{2yz+2y+2} \\ \frac{4z}{3z^2+4x^2+16}=\frac{4z}{(4x^2+z^2)+(2z^2+8)+8} \leq \frac{4z}{4zx+8z+8}=\frac{z}{zx+2z+2}[/tex]

[tex]\Rightarrow P \leq \frac{x}{2xy+2x+4}+\frac{y}{2yz+2y+2}+\frac{z}{zx+2z+2} \\ = \frac{x}{2xy+2x+2xyz}+\frac{y}{2yz+2y+2}+\frac{yz}{xyz+2yz+2y} \\ =\frac{1}{2y+2+2yz}+\frac{y}{2yz+2y+2}+\frac{yz}{2yz+2y+2}=2[/tex]

Vậy P max =2. Dấu bằng khi (x,y,z)=(1,1,2)