Toán C.2 :
Ta có : \(P=xy+yz+zx-xyz\Leftrightarrow2P=2.\left(xy+yz+zx\right)-2xyz\)
\(=2.\left(xy+yz+zx\right)+x^2+y^2+z^2-1\)
\(=\left(x+y+z\right)^2-1\)
Vì : \(x^2+y^2+z^2+2xyz=1\)
\(\Rightarrow z^2+2xyz=1-x^2-y^2\)
\(\Rightarrow z^2+2xyz+x^2y^2=1-x^2-y^2+x^2y^2\)
\(\Rightarrow\left(z+xy\right)^2=\left(1-x^2\right)\left(1-y^2\right)\le\left(\dfrac{2-x^2-y^2}{2}\right)^2\)
\(\Rightarrow z+xy\le\dfrac{2-x^2-y^2}{2}\Rightarrow z\le\dfrac{2-x^2-y^2-2xy}{2}=\dfrac{2-\left(x+y\right)^2}{2}\)
Có : \(\left(x+y-1\right)^2\ge0\)
\(\Leftrightarrow\left(x+y\right)^2-2.\left(x+y\right)+1\ge0\)
\(\Leftrightarrow x+y\le\dfrac{\left(x+y\right)^2+1}{2}\)
\(\Leftrightarrow x+y+z\le\dfrac{\left(x+y\right)^2+1}{2}+\dfrac{2-\left(x+y\right)^2}{2}=\dfrac{3}{2}\)
\(\Rightarrow\left(x+y+z\right)^2-1\le\dfrac{5}{4}\)
\(\Rightarrow2P\le\dfrac{5}{4}\Rightarrow P\le\dfrac{5}{8}\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z=\dfrac{1}{2}\)