sai đề

Theo nguyên lý Dirichlet, trong 3 số x;y;z luôn có 2 số cùng phía so với \(\dfrac{1}{2}\)

Không mất tính tổng quát, giả sử đó là y và z 

\(\Rightarrow\left(y-\dfrac{1}{2}\right)\left(z-\dfrac{1}{2}\right)\ge0\Leftrightarrow yz-\dfrac{1}{2}\left(y+z\right)+\dfrac{1}{4}\ge0\)

\(\Leftrightarrow y+z-yz\le\dfrac{1}{2}+yz\)

Mặt khác từ giả thiết:

\(1-x^2=y^2+z^2+2xyz\ge2yz+2xyz\)

\(\Leftrightarrow\left(1-x\right)\left(1+x\right)\ge2yz\left(1+x\right)\)

\(\Leftrightarrow1-x\ge2yz\)

\(\Rightarrow yz\le\dfrac{1-x}{2}\)

Do đó:

\(A=yz+x\left(y+z-yz\right)\le yz+x\left(\dfrac{1}{2}+yz\right)=\dfrac{1}{2}x+yz\left(x+1\right)\le\dfrac{1}{2}x+\left(\dfrac{1-x}{2}\right)\left(x+1\right)\)

\(\Rightarrow A\le-\dfrac{1}{2}x^2+\dfrac{1}{2}x+\dfrac{1}{2}=-\dfrac{1}{2}\left(x-\dfrac{1}{2}\right)^2+\dfrac{5}{8}\le\dfrac{5}{8}\)

\(A_{max}=\dfrac{5}{8}\) khi \(\left(x;y;z\right)=\left(\dfrac{1}{2};\dfrac{1}{2};\dfrac{1}{2}\right)\)

\(xy+yz+zx-xyz=1-x-y-z+xy+yz+zx-xyz\)

\(=\left(1-x\right)-y\left(1-x\right)-z\left(1-x\right)+yz\left(1-x\right)\)

\(=\left(1-x\right)\left(1-y-z+yz\right)=\left(1-x\right)\left(1-y\right)\left(1-z\right)\)

\(xy+yz+zx+xyz+2=1+x+y+z+xy+yz+zx+xyz\)

\(=\left(1+x\right)+y\left(1+x\right)+z\left(1+x\right)+yz\left(1+x\right)\)

\(=\left(1+x\right)\left(1+y\right)\left(1+z\right)\)

\(1+x+y+z=1+1\Rightarrow1+x=\left(1-y\right)+\left(1-z\right)\ge2\sqrt{\left(1-y\right)\left(1-z\right)}\)

Tương tự ta cũng có: \(1+y\ge2\sqrt{\left(1-z\right)\left(1-x\right)}\)

\(1+z\ge2\sqrt{\left(1-x\right)\left(1-y\right)}\)

Vậy \(S\le\frac{\left(1-x\right)\left(1-y\right)\left(1-z\right)}{8\left(1-x\right)\left(1-y\right)\left(1-z\right)}=\frac{1}{8}\)