x+y+z=0

=>(x+y+z)²=x²+y²+z²+2xy+2yz+2xz=0

<=>x²+y²+z²+2(xy+yz+xz)=0

<=>x²+y²+z²+2.0=0

<=>x²+y²+z²=0

<=>x=y=z=0( điều phải chứng minh )

Ta có 

x + y + z - xy - yz - xz \(\le1\)

\(\Leftrightarrow\left(1-x\right)+\left(xy-y\right)+\left(yz-xyz\right)+\left(xz-z\right)+xyz\ge0\)

\(\Leftrightarrow\left(1-x\right)\left(1-y-z+yz\right)+xyz\ge0\)

\(\Leftrightarrow\left(1-x\right)\left(\left(1-y\right)+\left(-z+yz\right)\right)+xyz\ge0\)

\(\Leftrightarrow\left(1-x\right)\left(1-y\right)\left(1-z\right)+xyz\ge0\)

Đúng vì theo đề ta có: \(\hept{\begin{cases}1-x\ge0\\1-y\ge0\\1-z\ge0\end{cases}}\)và \(\hept{\begin{cases}x\ge0\\y\ge0\\z\ge0\end{cases}}\)

Vậy ta có ĐPCM

Giải

Ta có : ( x + y + z )\(^2\)= x\(^2\)+ y\(^2\)+ z\(^2\)+ 2( xy + yz + zx )

Suy ra                      0 = x\(^2\)+ y\(^2\)+ z\(^2\)+ 2.0

hay                           0 = x\(^2\)+ y\(^2\)+ z\(^2\)

Vậy x = y = z ( = 0 )

Ta có: \(\frac{xy}{z}+\frac{yz}{x}\ge2\cdot\sqrt{\frac{xy}{z}\cdot\frac{yz}{x}}=2\cdot\sqrt{\frac{xy^2z}{xz}}=2\cdot y\)

\(\frac{yz}{x}+\frac{xz}{y}\ge2\cdot\sqrt{\frac{yz}{x}\cdot\frac{xz}{y}}=2\cdot\sqrt{z^2}=2z\)

\(\frac{xy}{z}+\frac{xz}{y}\ge2\cdot\sqrt{\frac{xy}{z}\cdot\frac{xz}{y}}=2\cdot\sqrt{x^2}=2x\)

Do đó: \(\left(\frac{xy}{z}+\frac{yz}{x}\right)+\left(\frac{yz}{x}+\frac{xz}{y}\right)+\left(\frac{xz}{y}+\frac{xy}{z}\right)\ge2\left(x+y+z\right)\)

=>\(2\left(\frac{xy}{z}+\frac{xz}{y}+\frac{yz}{x}\right)\ge2\left(x+y+z\right)\)

=>\(\frac{xy}{z}+\frac{xz}{y}+\frac{yz}{x}\ge\left(x+y+z\right)\)

Ta có :

\(\left(x+y+z\right)^2\)

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

\(\Rightarrow0=x^2+y^2+z^2+2.0\)

\(\Rightarrow0=x^2+y^2+z^2\)

Vậy \(x=y=z\left(=0\right)\)(đpcm)

ủa đây là toám lớp 1 hả anh

cauchy phần mẫu @@