Giải
P = \(\frac{x}{xy+3z}+\frac{y}{yz+3z}+\frac{z}{zx+3x}\)\(=\frac{x}{xy+\left(x+y+z\right)z}+\frac{y}{yz+\left(x+y+z\right)x}+\frac{z}{zx+\left(x+y+z\right)y}\)
\(=\frac{x}{\left(x+z\right)\left(y+z\right)}+\frac{y}{\left(x+y\right)\left(x+z\right)}+\frac{z}{\left(x+y\right)\left(y+z\right)}\)\(=\frac{x\left(x+y\right)+y\left(y+z\right)+z\left(z+x\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)=\(=\frac{x^2+y^2+z^2+xy+yz+zx}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
\(=\frac{\left(x+y+z\right)^2-\left(xy+yz+xz\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
Theo BĐT CÔSI : \(\left(x+y\right)\left(y+z\right)\left(z+x\right)\le\frac{\left(2x+2y+2z\right)^3}{27}=8\)
\(xy+yz+zx\le\frac{\left(x+y+z\right)^2}{3}=3\)
Do Đó : \(P\ge\frac{3^2-3}{8}=\frac{2}{3}\)
Vậy Min P= 2/3 dấu = <=> x=y=z=1
tik cho mik nha !!!!