\(P=xyz-(xy+2yz+zx)=yz(x-2)-x(y+z)\) \(\geq \left ( \frac{y+z}{z} \right )^2(x-2)-x(y+z)=\frac{(1-x)^2(x-2)}{4}-x(1-x)=f(x)\) do \(0\leq x\leq 1\) Xét hàm số: \(f(x)=\frac{(1-x)^2(x-2)}{4}-x(1-x),x\in \left [ 0;1 \right ]\) khảo sát hàm số \(y=f(x), x\in \left [ 0;1 \right ]\) , ta có \(f(x)\geq f(0)=-\frac{1}{2},\forall x\in \left [ 0;1 \right ]\) Vậy \(P\geq -\frac{1}{2}\). Mặt khác, khi \(x=0, y=z=\frac{1}{2}\) thì \(P=-\frac{1}{2}\). Vậy \(min \ \ P=\frac{1}{2}\) Ta có \(P=xyz-(xy+2yz+zx)\leq yz-(xy+2yz+zx)\)\(-(xy+yz+zx)\leq 0\) Khi x = 1, y = z =0 \(\Rightarrow\) P =0. Vậy max P = 0