\(P=\dfrac{xy}{1+x+y}+\dfrac{yz}{1+y+z}+\dfrac{xz}{1+z+x}\)
\(P+3=\dfrac{xy}{1+x+y}+1+\dfrac{yz}{1+y+z}+1+\dfrac{xz}{1+z+x}+1\)
\(P+3=\dfrac{\left(x+1\right)\left(y+1\right)}{1+x+y}+\dfrac{\left(y+1\right)\left(z+1\right)}{1+y+z}+\dfrac{\left(x+1\right)\left(z+1\right)}{1+z+x}\)
\(P+3=\dfrac{\left(x+1\right)\left(y+1\right)\left(z+1\right)}{\left(1+x+y\right)\left(z+1\right)}+\dfrac{\left(x+1\right)\left(y+1\right)\left(z+1\right)}{\left(x+1\right)\left(1+y+z\right)}+\dfrac{\left(x+1\right)\left(y+1\right)\left(z+1\right)}{\left(y+1\right)\left(1+z+x\right)}\)
\(P+3=\left(x+1\right)\left(y+1\right)\left(z+1\right)\left[\dfrac{1}{\left(1+x+y\right)\left(z+1\right)}+\dfrac{1}{\left(x+1\right)\left(1+y+z\right)}+\dfrac{1}{\left(y+1\right)\left(1+z+x\right)}\right]\)
\(\ge\left(x+1\right)\left(y+1\right)\left(z+1\right)\cdot\dfrac{9}{\left(1+x+y\right)\left(z+1\right)+\left(x+1\right)\left(1+y+z\right)+\left(y+1\right)\left(1+z+x\right)}\)
\(=\left(x+1\right)\left(y+1\right)\left(z+1\right)\cdot\dfrac{9}{\text{ }2xy+2yz+2xz+3x+3y+3z+3}\)
\(=\left(x+1\right)\left(y+1\right)\left(z+1\right)\cdot\dfrac{9}{\text{ }2xy+2yz+2xz+3\cdot2xyz}\)
\(=\left(x+1\right)\left(y+1\right)\left(z+1\right)\cdot\dfrac{9}{\text{ }2\left(xy+yz+xz+3xyz\right)}\)
Lại có:
\(\left(x+1\right)\left(y+1\right)\left(z+1\right)=xyz+xy+yz+xz+x+y+z+1\)
\(=xyz+xy+yz+xz+2xyz=xy+yz+xz+3xyz\)
\(\Rightarrow P+3\ge\left(xy+yz+xz+3xyz\right)\cdot\dfrac{9}{2\left(xy+yz+xz+3xyz\right)}\)
\(\Rightarrow P+3\ge\dfrac{9}{2}\Rightarrow P\ge\dfrac{9}{2}-3=\dfrac{3}{2}\)
Đẳng thức xảy ra khi \(x=y=z=\dfrac{1+\sqrt{3}}{2}\)
