Lời giải:
Từ đkđb suy ra:
$x-y=\frac{1}{z}-\frac{1}{y}=\frac{y-z}{yz}$

$y-z=\frac{1}{x}-\frac{1}{z}=\frac{z-x}{xz}$

$z-x=\frac{1}{y}-\frac{1}{x}=\frac{x-y}{xy}$

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

$\Leftrightarrow (x-y)(y-z)(z-x)(1-\frac{1}{x^2y^2z^2})=0$

$\Rightarrow (x-y)(y-z)(z-x)=0$ hoặc $1-\frac{1}{x^2y^2z^2}=1$

$\Rightarrow (x-y)(y-z)(z-x)=0$ hoặc $x^2y^2z^2=1$
Nếu $(x-y)(y-z)(z-x)=0$

$\Rightarrow x=y$ hoặc $y=z$ hoặc $z=x$

Không mất tquat giả sử $x=y$. Khi đó: $\frac{1}{y}=\frac{1}{z}$

$\Rightarrow y=z$

$\Rightarrow x=y=z$. Tương tự khi xét $y=z$ hoặc $z=x$ thì ta cũng thu được $x=y=z$
Vậy $x=y=z$ hoặc $x^2y^2z^2=1$

\(x+y+z=xyz\Leftrightarrow\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}=1\)

\(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}=\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2-2\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}\right)=2^2-2.1=2\) (đpcm)

Đề sai nhé, \(\dfrac{z^2}{x+1}\) mới đúng nha

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

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

Dấu "=" xảy ra \(\Leftrightarrow x=y=z=1\)

Ta có: \(x+y+z\ge3\sqrt[3]{xyz}=3\)

\(\Rightarrow x+y+z+3\le2\left(x+y+z\right)\)

\(\Rightarrow\left(x+y+z\right)^2\ge\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2\ge3\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}\right)=\dfrac{3\left(x+y+z\right)}{xyz}\Rightarrow x+y+z\ge\dfrac{3}{xyz}\)

\(x+y+z=\dfrac{x+y+z}{3}+\dfrac{2\left(x+y+z\right)}{3}\ge\dfrac{1}{3}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)+\dfrac{2}{3}.\dfrac{3}{xyz}\ge\dfrac{1}{3}\left(\dfrac{9}{x+y+z}\right)+\dfrac{2}{xyz}=\dfrac{3}{x+y+z}+\dfrac{2}{xyz}\left(đpcm\right)\)

\(dấu"="xảy\) \(ra\Leftrightarrow x=y=z=1\)

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 

Giả thiết thiếu rồi em, chỗ \(\dfrac{1}{x+1}+...\) thiếu đoạn sau nữa