VT=\(x^3+y^3+z^3-3xyz=\left(x+y\right)^3+z^3-3x^2y-3xy^2-3xyz\)

\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy.\left(x+y+z\right)\)

\(=\left(x+y\right)^2-\left(x+y\right).z+z^2-3xy\left(\text{vì }x+y+z=1\right)\)

\(=x^2+2xy+y^2-xz-yz+z^3-3xy\)

\(=x^2+y^2+z^2-xy-yz-xz\)

\(=\frac{1}{2}.\left(2x^2+2y^2+2z^2-2xy-2yz-2xz\right)\)

\(=\frac{1}{2}.\left[\left(x^2-2xy-y^2\right)+\left(y^2-2yz+z^2\right)+\left(x^2-2xz+z^2\right)\right]\)

\(=\frac{1}{2}.\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]\)=VP

=>dpcm

Ta có : \(x^3+y^3+z^3-3xyz=\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz\)

\(=x+y+z\left(x^2+y^2+z^2+2xy+xz+yz\right)-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)

\(=x^2+y^2+z^2-xy-yz-xz=\frac{\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2xz+x^2\right)}{2}=\frac{1}{2}\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]\)

Đặt \(\left(\frac{1}{x};\frac{1}{y};\frac{1}{z}\right)=\left(a;b;c\right)\Rightarrow ab+bc+ca=3\). Tìm Min:\(P=\Sigma_{cyc}\frac{a^3}{\left(b+2c\right)}\)

Auto làm nốt:3

\(\frac{1}{6}\)nha bạn

Áp dụng bất đẳng thức Cauchy - Schwarz 

\(\Rightarrow\hept{\begin{cases}\frac{x^3}{\left(2x+y\right)\left(y+z\right)}+\frac{2x+y}{8}+\frac{y+z}{8}\ge3\sqrt[3]{\frac{x^3}{64}}=\frac{3x}{4}\\\frac{y^3}{\left(2y+z\right)\left(z+x\right)}+\frac{2y+z}{8}+\frac{x+z}{8}\ge3\sqrt[3]{\frac{y^3}{64}}=\frac{3y}{4}\\\frac{z^3}{\left(2z+x\right)\left(x+y\right)}+\frac{2z+x}{8}+\frac{x+y}{8}\ge3\sqrt[3]{\frac{z^3}{64}}=\frac{3z}{4}\end{cases}}\)

\(\Rightarrow\frac{x^3}{\left(2x+y\right)\left(y+z\right)}+\frac{y^3}{\left(2y+z\right)\left(x+z\right)}+\frac{z^3}{\left(2z+x\right)\left(x+y\right)}+\frac{5\left(x+y+z\right)}{8}\ge\frac{3\left(x+y+z\right)}{4}\)

\(\Rightarrow\frac{x^3}{\left(2x+y\right)\left(y+z\right)}+\frac{y^3}{\left(2y+z\right)\left(x+z\right)}+\frac{z^3}{\left(2z+x\right)\left(x+y\right)}+\frac{5}{8}\ge\frac{3}{4}\)

\(\Rightarrow\frac{x^3}{\left(2x+y\right)\left(y+z\right)}+\frac{y^3}{\left(2y+z\right)\left(x+z\right)}+\frac{z^3}{\left(2z+x\right)\left(x+y\right)}\ge\frac{1}{8}\)

\(\Leftrightarrow P_{min}=\frac{1}{8}\)

Không mất tính tổng quát giả sử \(z=min\left(x;y;z\right)\)

Từ giả thiết x+y+z=3 => \(3z\le x+y+z\)Do đó \(0\le z\le1\)

Đặt x=1+a; y=1+b; c=1-a-b. Do 0 =<c=<1 nên 0 =< a+b =< 1

Ta có \(\left(x-1\right)^3+\left(y-1\right)^3+\left(z-1\right)^3=a^3+b^3+\left(-a-b\right)^3=-3ab\left(a+b\right)\)

Mặt khác \(\left(a-b\right)^2\ge0\forall a,b\Rightarrow ab\le\frac{\left(a+b\right)^2}{4}\)

\(\Rightarrow ab\left(a+b\right)\le\frac{\left(a+b\right)^2}{4}\le\frac{1}{4}\left(0\le a+b\le1\right)\)

\(\Rightarrow-3ab\left(a+b\right)\ge\frac{-3}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=\frac{1}{2}\)

Khi đó \(x=y=\frac{3}{2};z=0\)

phân tích gt sau đó suy ra x+y+x=0 

từ đây tính đc x+y=? y+z=? x+z=? 

ta được kết quả là'; -2006

Xét \(x^3+y^3+z^3=3xyz\)

\(x^3+y^3+z^3-3xyz=0\)

\(\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz=0\)

\(\left(x+y\right)^3+z^3-3xy\left(x+y+z\right)=0\)

\(\left(x+y+z\right)\left(x^2+2xy+y^2-xy-yz+z^2\right)-3xy\left(x+y+z\right)=0\)

\(\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)=0\)

TH1:\(x+y+z=0\) 

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

Thay (1) vô pt cần tính:

\(\frac{2016xyz}{-z.-x.-y}=\frac{2016xyz}{-\left(xyz\right)}=-2016\)

TH2:\(x^2+y^2+z^2-xy-yz-xz=0\)

Nhân 2 vế với 2

\(2x^2+2y^2+2z^2-2xy-2yz-2xz=0\)

\(x^2-2xy+y^2+x^2-2xz+z^2+y^2-2yz+z^2=0\)

\(\left(x-y\right)^2+\left(x-z\right)^2+\left(y-z\right)^2=0\)

Do VT dương

\(\Rightarrow\hept{\begin{cases}\left(x-y\right)^2=0\\\left(x-z\right)^2=0\\\left(y-z\right)^2=0\end{cases}\Rightarrow}\hept{\begin{cases}x-y=0\\x-z=0\\y-z=0\end{cases}\Rightarrow}\hept{\begin{cases}x=y\\x=z\\y=z\end{cases}}\Rightarrow x=y=z\)

Thay y,z ở pt cần tính là x

\(\Rightarrow\frac{2016x.x.x}{\left(x+x\right)\left(x+x\right)\left(x+x\right)}=\frac{2016x^3}{2x.2x.2x}=\frac{2016x^3}{8x^3}=\frac{2016}{8}=252\)

Vậy pt có thể = -2016 khi x + y + z = 0

       pt có thể = 252 khi \(x^2+y^2+z^2-xy-xz-yz=0\)

\(x^3+y^3+z^3=3xyz\)

\(x^3+y^3+z^3-3xyz=0\)

\(x^3+3x^2y+3xy^2+y^3+z^3-3x^2y-3xy^2-3xyz=0\)

\(\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz=0\)

\(\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)=0\)

\(\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)=0\)

\(\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+y+z=0\\x^2+y^2+z^2-xy-xz-yz=0\end{cases}}\)

TH1 : \(x+y+z=0\)

\(\Rightarrow\hept{\begin{cases}x+y=-z\\x+z=-y\\y-z=-x\end{cases}}\)

Thay vào biểu thức ta có :

\(\frac{2016xyz}{\left(-z\right)\left(-x\right)\left(-y\right)}=\frac{2016xyz}{-xyz}=-2016\)

TH2 : \(x^2+y^2+z^2-xy-xz-yz=0\)

\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2xz-2yz=0\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(x^2-2xz+z^2\right)+\left(y^2-2yz+z^2\right)=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(x-z\right)^2+\left(y-z\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}x-y=0\\x-z=0\\y-z=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=y\\x=z\\y=z\end{cases}\Rightarrow}x=y=z}\)

Từ đây ta biến đổi biểu thức :

\(\frac{2016x^3}{2x\cdot2x\cdot2x}=\frac{2016x^3}{8x^3}=252\)

Vậy........