\(x+\frac{1}{y}=y+\frac{1}{z}=z+\frac{1}{x}\)\(\Rightarrow\hept{\begin{cases}x-y=\frac{1}{z}-\frac{1}{y}=\frac{y-z}{xy}\\y-z=\frac{1}{x}-\frac{1}{z}=\frac{z-x}{xz}\\z-x=\frac{1}{y}-\frac{1}{x}=\frac{x-y}{xy}\end{cases}}\)

\(\Rightarrow\left(x-y\right)\left(y-z\right)\left(z-x\right)=\frac{\left(x-y\right)\left(y-z\right)\left(z-x\right)}{\left(xyz\right)^2}\)

\(\Leftrightarrow\frac{1}{\left(xyz\right)^2}=1\Rightarrow xyz=\pm1\)(đpcm)

Từ giả thiết , ta có :

\(xyz=\left(1-x\right)\left(1-y\right)\left(1-z\right)\left(1\right)\)

\(\Rightarrow1=\left(\frac{1}{x}-1\right)\left(\frac{1}{y}-1\right)\left(\frac{1}{z}-1\right)\)

Áp dụng bất đẳng thức sau : \(abc\le\left(\frac{a+b+c}{3}\right)^3\) ta có :

\(1=\left(\frac{1}{x}-1\right)\left(\frac{1}{y}-1\right)\left(\frac{1}{z}-1\right)\le\left(\frac{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-3}{3}\right)^3\)

\(\Rightarrow3\le\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-3\)

\(\Rightarrow6\le\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)

\(\Rightarrow6xyz\le xy+yz+zx\left(2\right)\)

Từ ( 1 ) và ( 2 ) suy ra:

\(3-3\left(x+y+z\right)+3\left(xy+yz+zx\right)=6xyz\le xy+yz+zx\)

\(\Rightarrow0\ge3-3\left(x+y+z\right)+2\left(xy+yz+zx\right)\)

Cộng 2 vế của bất đẳng thức trên cho \(\left(x^2+y^2+z^2\right)\)ta được:

\(x^2+y^2+z^2\ge\left(x+y+z\right)^2-3\left(x+y+z+3\right)=\left(x+y+z-\frac{3}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Dấu '' = '' xảy ra khi và chỉ khi \(x=y=z=\frac{1}{2}\) 

ta có:

xyz=(1-x).(1-y).(1-z)                                 (1)

=>1=(1:x-1).(1:y-1).(1:z-1)

\(x,y,z>0\)

Áp dụng BĐT Caushy cho 3 số ta có:

\(x^3+y^3+z^3\ge3\sqrt[3]{x^3y^3z^3}=3xyz\ge3.1=3\)

\(P=\dfrac{x^3-1}{x^2+y+z}+\dfrac{y^3-1}{x+y^2+z}+\dfrac{z^3-1}{x+y+z^2}\)

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

Áp dụng BĐT Caushy-Schwarz ta có:

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

\(\ge\dfrac{\left(3-3\right)^2}{\left(x^2+y+z\right)\left(x^3-1\right)+\left(x+y^2+z\right)\left(y^3-1\right)+\left(x+y^2+z\right)\left(y^3-1\right)}=0\)

\(P=0\Leftrightarrow x=y=z=1\)

Vậy \(P_{min}=0\)

Ta có \(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+\dfrac{2}{xyz}=1\)

\(\Leftrightarrow\dfrac{\left(yz\right)^2+\left(xz\right)^2+\left(xy\right)^2+2xyz}{\left(xyz\right)^2}=1\)

<=> (xy)² + (yz)² + (zx)² + 2xyz = (xyz)² 

<=> (xy)² + (yz)² + (xz)² + 2xyz(x + y + z) = (xyz)² 

<=> (xy + yz + zx)² = (xyz)² 

<=> \(\left[{}\begin{matrix}xy+yz+zx=xyz\\xy+yz+zx=-xyz\end{matrix}\right.\)

+) Khi xy + yz + zx = -xyz 

=> \(\dfrac{xy+yz+zx}{xyz}=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=-1< 0\left(\text{loại}\right)\)

=> xy + yz + zx = xyz 

<=> \(xyz\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=xyz\Leftrightarrow xyz\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}-1\right)=0\)

<=> \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1\)

<=> \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{x+y+z}\)

<=> \(\dfrac{x+y}{xy}=\dfrac{-\left(x+y\right)}{\left(x+y+z\right)z}\)

<=> \(\left(x+y\right)\left(\dfrac{1}{xz+yz+z^2}+\dfrac{1}{xy}\right)=0\)

<=> \(\dfrac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{\left(zx+yz+z^2\right)xy}=0\)

<=> \(\left[{}\begin{matrix}x=-y\\y=-z\\z=-x\end{matrix}\right.\)

Khi x = -y => y = 1 => P = 1

Tương tự y = -z ; z = -x được P = 1

Vậy P = 1 

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

=>2(xy+yz+xz)=0

=>xy+xz+yz=0

=>xy/xyz+xz/xyz+yz/xyz=0

=>1/x+1/y+1/z=0