Lời giải:
Áp dụng TCDTSBN:

$\frac{1}{x+y+z}=\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{y+z+1+x+z+2+x+y-3}{x+y+z}=\frac{2(x+y+z)}{x+y+z}=2$

\(\Rightarrow \left\{\begin{matrix} x+y+z=\frac{1}{2}\\ y+z+1=2x\\ x+z+2=2y\\ x+y-3=2z\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x+y+z=\frac{1}{2}\\ x+y+z+1=3x\\ x+y+z+2=3y\\ x+y+z-3=3z\end{matrix}\right.\)

\(\left\{\begin{matrix} \frac{1}{2}+1=3x\\ \frac{1}{2}+2=3y\\ \frac{1}{2}-3=3z\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x=\frac{1}{2}\\ y=\frac{5}{6}\\ z=\frac{-5}{6}\end{matrix}\right.\)

Áp dụng t/c dtsbn ta có:

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

\(\dfrac{1}{x+y+z}=2\Rightarrow2x+2y+2z=1\Rightarrow x+y+z=0,5\Rightarrow\left\{{}\begin{matrix}x+y=0,5-z\\y+z=0,5-x\\x+z=0,5-y\end{matrix}\right.\\ \dfrac{y+z+1}{x}=2\Rightarrow y+z+1=2x\Rightarrow0,5-x+1=2x\Rightarrow x=0,5\\ \dfrac{x+z+2}{y}=2\Rightarrow x+z+2=2y\Rightarrow0,5-y+2=2y\Rightarrow y=\dfrac{5}{6}\\ \dfrac{x+y-3}{z}=2\Rightarrow x+y-3=2z\Rightarrow0,5-z-3=2z\Rightarrow z=-\dfrac{5}{6}\)

Lời giải:

Bạn cần bổ sung điều kiện $x,y,z>0$

\(P=\frac{1}{x.\frac{y^2+z^2}{y^2z^2}}+\frac{1}{y.\frac{z^2+x^2}{z^2x^2}}+\frac{1}{z.\frac{x^2+y^2}{x^2y^2}}=\frac{1}{x(\frac{1}{y^2}+\frac{1}{z^2})}+\frac{1}{y(\frac{1}{z^2}+\frac{1}{x^2})}+\frac{1}{z(\frac{1}{x^2}+\frac{1}{y^2})}\)

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

Vì $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=3\Rightarrow x^2, y^2, z^2>\frac{1}{3}$

Xét hiệu:

\(\frac{x}{3x^2-1}-\frac{1}{2x^2}=\frac{(x-1)^2(2x+1)}{2x^2(3x^2-1)}\geq 0\) với mọi $x>0$ và $x^2>\frac{1}{3}$

$\Rightarrow \frac{x}{3x^2-1}\geq \frac{1}{2x^2}$

Hoàn toàn tương tự với các phân thức còn lại và cộng theo vế ta có:

$P\geq \frac{1}{2}(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2})=\frac{3}{2}$

Vậy $P_{\min}=\frac{3}{2}$ khi $x=y=z=1$

Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{y+z+1}=\dfrac{y}{x+z+1}=\dfrac{z}{x+y-2}=x+y+z=\dfrac{x+y+z}{y+z+1+x+z+1+x+y-2}=\dfrac{x+y+z}{2x+2y+2z}=\dfrac{x+y+z}{2\left(x+y+z\right)}=\dfrac{1}{2}\)

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

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

\(\dfrac{z}{x+y-2}=\dfrac{1}{2}\Rightarrow x+y-2=2z\)

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

Thay (1) vào (3) ta có:

\(x+y+z=\dfrac{1}{2}\\ \Rightarrow x+2x-1=\dfrac{1}{2}\\ \Rightarrow3x=\dfrac{3}{2}\\ \Rightarrow x=\dfrac{1}{2}\)

Thay (2) vào (3) ta có:

\(x+y+z=\dfrac{1}{2}\\ \Rightarrow y+2y-1=\dfrac{1}{2}\\ \Rightarrow3y=\dfrac{3}{2}\\ \Rightarrow y=\dfrac{1}{2}\)

Ta có:

\(x+y+z=\dfrac{1}{2}\\ \Rightarrow\dfrac{1}{2}+\dfrac{1}{2}+z=\dfrac{1}{2}\\ \Rightarrow z=-\dfrac{1}{2}\)

TH1: \(x+y+z=0\Rightarrow x=y=z=0\)

TH2: \(x+y+z\ne0\)

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}2x+2y+2z=1\\y+z=0\\x+z=0\\x+y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2.1+2z=1\\y=-z\\x=-z\\x+y=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}z=-\dfrac{1}{2}\\x=\dfrac{1}{2}\\y=\dfrac{1}{2}\\\end{matrix}\right.\)

Vậy \(\left(x;y;z\right)=\left(0;0;0\right);\left(\dfrac{1}{2};\dfrac{1}{2};-\dfrac{1}{2}\right)\)

\(\dfrac{x}{x^2+yz}+\dfrac{y}{y^2+zx}+\dfrac{z}{z^2+xy}\le\dfrac{x}{2\sqrt{x^2yz}}+\dfrac{y}{2\sqrt{y^2zx}}+\dfrac{z}{2\sqrt{z^2xy}}=\dfrac{1}{2}\left(\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}+\dfrac{1}{\sqrt{xy}}\right)\le\dfrac{1}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{3}{2}\).

Đẳng thức xảy ra khi x = y = z = 1.

Lời giải:
Áp dụng BĐT Cauchy-Schwarz:

$3=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{9}{x+y+z}$
$\Rightarrow x+y+z\geq 3$

Áp dụng BĐT AM-GM:

$\frac{y^2}{2}+\frac{1}{2}\geq y$

$\frac{z^3}{3}+\frac{1}{3}+\frac{1}{3}\geq z$

$\Rightarrow P+\frac{7}{6}\geq x+y+z=3$

$\Rightarrow P\geq \frac{11}{6}$

Giá trị này đạt tại $x=y=z=1$

\(\Rightarrow\dfrac{z+y+1}{x}=\dfrac{x+z+1}{y}=\dfrac{x+y-2}{z}=\dfrac{2\left(x+y+z\right)}{x+y+z}=2=x+y+z\\ \Rightarrow\left\{{}\begin{matrix}z+y+1=2x\\x+z+1=2y\\x+y-2=2z\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y+z=2x-1\\x+z=2y-1\\x+y=2z+2\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}2x-1=2-x\\2y-1=2-y\\2z+2=2-z\end{matrix}\right.\Rightarrow\left(x,y,z\right)=\left(1;1;0\right)\)

Lời giải:

Áp dụng BĐT Cô-si:

\(x^2+y^2+z^2\geq \frac{(x+y+z)^2}{3}\)

\(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\geq \frac{1}{3}(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2\geq \frac{1}{3}.(\frac{9}{x+y+z})^2=\frac{27}{(x+y+z)^2}\)

\(\Rightarrow P\geq \frac{(x+y+z)^2}{3}+\frac{27}{(x+y+z)^2}\)

Áp dụng BĐT Cô-si:

\(\frac{(x+y+z)^2}{3}+\frac{1}{3(x+y+z)^2}\geq \frac{2}{3}\)

\(\frac{80}{3(x+y+z)^2}\geq \frac{80}{3}\)

\(\Rightarrow P\geq \frac{2}{3}+\frac{80}{3}=\frac{82}{3}\)

Vậy $P_{\min}=\frac{82}{3}$ khi $x=y=z=\frac{1}{3}$