\(S=x\left(3x+2y+z\right)+\left(y-x\right)\left(2y+z\right)+\left(z-y\right).y\)

\(S\le4x+3\left(y-x\right)+z-y=x+2y+z\)

\(S\le\dfrac{1}{3}\left(3x+2y+z\right)+\dfrac{2}{3}\left(2y+z\right)\le\dfrac{1}{3}.4+\dfrac{2}{3}.3=\dfrac{10}{3}\)

Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(\dfrac{1}{3};1;1\right)\)

Khai triển Abel ta có:

\(S=\left(z-y\right)z+\left(y-x\right)\left(z+2y\right)+x\left(3x+2y+z\right)\)

\(\le\left(z-y\right).1+\left(y-x\right).3+4x=x+2y+z\)

\(=\left(1-1\right)z+\left(1-\dfrac{1}{3}\right)\left(2y+z\right)+\dfrac{1}{3}\left(3x+2y+z\right)\)

\(\le\dfrac{2}{3}.3+\dfrac{1}{3}.4=\dfrac{10}{3}\)

Dấu = xảy ra khi \(x=\dfrac{1}{3},y=z=1\)

trả lời được tớ cho3 k

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\(\Leftrightarrow\left\{{}\begin{matrix}\left(3x-2y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2\ge0\\\left(3x-2y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2\le1\end{matrix}\right.\) \(\begin{matrix}\left(1\right)\\\left(2\right)\end{matrix}\)

(1) đúng với mọi x,y,z thuộc R =>đúng với mọi x,y,z thuộcZ

có

điều kiện cần thỏa mãn (2)

\(\left\{{}\begin{matrix}\left|3x-2y\right|\le1\\\left|y+z\right|\le1\\\left|z-x\right|\le1\end{matrix}\right.\) \(\begin{matrix}\left(a\right)\\\left(b\right)\\\left(c\right)\end{matrix}\)

\(\left(b\right)+\left(c\right)\Leftrightarrow\left|y+z\right|+\left|z-x\right|=\left|y+z\right|+\left|x-z\right|\ge\left|y+z+x-z\right|=\left|y+x\right|\) (d)

\(\left|3x-2y\right|+\left|2y+2x\right|\ge\left|3x-2y+2y+2x\right|=\left|5x\right|\)

cần : \(\left|5x\right|\le2\Leftrightarrow x=\left\{0;\pm1\right\}\)

x=0 từ (a) => y =0 ; từ (b) (c)=z =0 ; (x;y;z) =(0;0;0)

x=1 từ (a) =y={1;2}

với y=1 từ (b) => z=-1 ; (x;y;z) =(1;1;-1)

với y=2 từ (b) => z =-2 từ (c) $|-2-1| \ne 0$ loại

x=-1 từ (a) =y={-1;-2}

với y=-1 từ (b) => z= 1 ; (x;y;z) =(-1;-1;1)

với y=-2 từ (b) => z = 2 từ (c) $| 2+1| \ne 0$ loại

kết luận

(x;y;z) =(0;0;0);(1;1;1); (-1;-1;1)