Lời giải:
Nếu $x+y+z=0$ thì:

$\frac{x+y-z}{z}=\frac{-z-z}{z}=-2$

$\frac{y+z-x}{x}=\frac{-x-x}{x}=-2$

$\frac{z+x-y}{y}=\frac{-y-y}{y}=-2$ 

(thỏa mãn đkđb)

Khi đó:

$P=(1+\frac{x}{y})(1+\frac{y}{z})(1+\frac{z}{x})=\frac{(x+y)(y+z)(z+x)}{xyz}$

$=\frac{(-z)(-x)(-y)}{xyz}=\frac{-xyz}{xyz}=-1$

Nếu $x+y+z\neq 0$

Áp dụng TCDTSBN:

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

$\Rightarrow x+y=2z; y+z=2x, z+x=2y$. Khi đó:

$P=\frac{(x+y)(y+z)(z+x)}{xyz}=\frac{2z.2x.2y}{xyz}=8$

Phân số cuối cùng chắc em ghi nhầm

\(\dfrac{x}{y+z+t}+\dfrac{y+z+t}{9x}\ge2\sqrt{\dfrac{x\left(y+z+t\right)}{9x\left(y+z+t\right)}}=\dfrac{2}{3}\)

Tương tự:

\(\dfrac{y}{z+t+x}+\dfrac{z+t+x}{9y}\ge\dfrac{2}{3}\)

\(\dfrac{z}{t+x+y}+\dfrac{t+x+y}{9z}\ge\dfrac{2}{3}\)

\(\dfrac{t}{x+y+z}+\dfrac{x+y+z}{9t}\ge\dfrac{2}{3}\)

Đồng thời:

\(\dfrac{8}{9}\left(\dfrac{y+z+t}{x}+\dfrac{z+t+x}{y}+\dfrac{t+x+y}{z}+\dfrac{x+y+z}{t}\right)\)

\(\ge\dfrac{8}{9}\left(\dfrac{3\sqrt[3]{yzt}}{x}+\dfrac{3\sqrt[3]{ztx}}{y}+\dfrac{3\sqrt[3]{txy}}{z}+\dfrac{3\sqrt[3]{xyz}}{t}\right)\)

\(\ge\dfrac{8}{3}.4\sqrt[4]{\dfrac{\sqrt[3]{yzt}.\sqrt[3]{ztx}.\sqrt[3]{txy}.\sqrt[3]{xyz}}{xyzt}}=\dfrac{32}{3}\)

Cộng vế:

\(VT\ge4.\dfrac{2}{3}+\dfrac{32}{3}=\dfrac{40}{3}\)

Dấu "=" xảy ra khi \(x=y=z=t\)

a)\(\dfrac{x}{y}+\dfrac{y}{x}-2=\dfrac{x^2+y^2-2xy}{xy}=\dfrac{\left(x-y\right)^2}{xy}\)\(\ge0\)

Vậy \(\dfrac{x}{y}+\dfrac{y}{x}\ge2\)

b) ta có: A=\(\left(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\right)-\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\)=\(\left(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\right)-2\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\)

A\(\ge\)\(\left(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\right)-2\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+2\)

=\(\left(\dfrac{x}{y}-1\right)^2+\left(\dfrac{y}{x}-1\right)^2\ge0\)

c) Từ câu b suy ra:

\(\left(\dfrac{x^4}{y^4}+\dfrac{y^4}{x^4}\right)-\left(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\right)\ge0\)

Vì \(\dfrac{x}{y}+\dfrac{y}{x}\ge2\)(câu a)

Nên:

\(\left(\dfrac{x^4}{y^4}+\dfrac{y^4}{x^4}\right)-\left(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\right)+\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\ge2\)

Đặt \(\left(x;y\right)=\left(\dfrac{1}{a};\dfrac{1}{b}\right)\)

BĐT trở thành: \(\dfrac{a^2}{b}+\dfrac{b^2}{a}+\dfrac{16ab}{a+b}\ge5\left(a+b\right)\)

\(\Leftrightarrow\dfrac{a^3+b^3}{ab}+\dfrac{16ab}{a+b}-5\left(a+b\right)\ge0\)

\(\Leftrightarrow\dfrac{\left(a+b\right)\left(a^3+b^3\right)+16a^2b^2-5ab\left(a+b\right)^2}{ab\left(a+b\right)}\ge0\)

\(\Leftrightarrow\dfrac{\left(a-b\right)^4}{ab\left(a+b\right)}\ge0\) (luôn đúng)

\(TH_1:x+y+z=0\Rightarrow\left\{{}\begin{matrix}x+y=-z\\y+z=-x\\x+z=-y\end{matrix}\right.\\ \Rightarrow Q=\dfrac{-z}{z}+\dfrac{-x}{x}+\dfrac{-y}{y}=-3\\ TH_2:x+y+z\ne0\\ \Rightarrow\dfrac{3x-2y+z}{x}=\dfrac{3y-2z+x}{y}=\dfrac{3z-2x+y}{z}=\dfrac{2x+2y+2z}{x+y+z}=2\\ \Rightarrow\left\{{}\begin{matrix}3x-2y+z=x\\3y-2z+x=y\\3z-2x+y=z\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x-2y=-z\\2y-2z=-x\\2z-2x=-y\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x-y=-\dfrac{z}{2}\\y-z=-\dfrac{x}{2}\\z-x=-\dfrac{y}{2}\end{matrix}\right.\)

\(\Rightarrow Q=-\dfrac{z}{2}:z-\dfrac{x}{2}:x-\dfrac{y}{2}:y=-\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{1}{2}=-\dfrac{3}{2}\)

b/ \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)

\(\Rightarrow\left(\dfrac{a}{b}\right)^3=\dfrac{a}{d}\left(1\right)\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\)

=> \(\left(\dfrac{a}{b}\right)^3=\left(\dfrac{a+b+c}{c+d+b}\right)^3\) (2)Từ (1) và (2)=>đpcm

Đặt \(\dfrac{x-y}{z}=m,\dfrac{y-z}{x}=n,\dfrac{z-x}{y}=p\), ta có:

\(\left(m+n+p\right)\left(\dfrac{1}{m}+\dfrac{1}{n}+\dfrac{1}{p}\right)=3+\dfrac{n+p}{m}+\dfrac{p+m}{n}+\dfrac{m+n}{p}\)

Tính \(\dfrac{n+p}{m}\) theo x, y, z ta được:

\(\dfrac{n+p}{m}=\dfrac{z}{x-y}.\dfrac{y^2-yz+xz-x^2}{xy}=\dfrac{z}{xy}\left(-x-y+x\right)\)

           \(=\dfrac{z}{xy}\left(-x-y-z+2z\right)=\dfrac{2x^2}{xy}\) vì \(\left(x+y+z\right)=0\)

Tương tự:    \(\dfrac{m+p}{n}=\dfrac{2x^2}{yz}.\dfrac{m+n}{p}=\dfrac{2y^2}{xz}\)

Vậy \(\left(m+n+p\right)\left(\dfrac{1}{m}+\dfrac{1}{n}+\dfrac{1}{p}\right)=3+\dfrac{2\left(x^3+y^3+z^3\right)}{xyz}=3+\dfrac{2.3xyz}{xyz}=3+6=9\)

TH1: \(x+y+z+t\ne0\) 

Áp dụng t/c dtsbn ta có:

\(\dfrac{x}{y+z+t}=\dfrac{y}{z+t+x}=\dfrac{z}{t+x+y}=\dfrac{t}{x+y+z}=\dfrac{x+y+z+t}{3\left(x+y+z+t\right)}=\dfrac{1}{3}\)\(\dfrac{x}{y+z+t}=\dfrac{1}{3}\Rightarrow3x=y+z+t\Rightarrow4x=x+y+z+t\\ \dfrac{y}{z+t+x}=\dfrac{1}{3}\Rightarrow3y=x+z+t\Rightarrow4y=x+y+z+t\\ \dfrac{z}{t+x+y}=\dfrac{1}{3}\Rightarrow3z=x+y+t\Rightarrow4z=x+y+z+t\\ \dfrac{t}{x+y+z}=\dfrac{1}{3}\Rightarrow3t=x+y+z\Rightarrow4t=x+y+z+t\)
\(\Rightarrow4x=4y=4z=4t\\ \Rightarrow x=y=z=t\)

\(P=\dfrac{x+y}{z+t}+\dfrac{y+z}{t+x}+\dfrac{z+t}{x+y}+\dfrac{t+x}{y+z}\\ =1+1+1+1\\ =4\)

TH1: \(x+y+z+t=0\) 

\(\Rightarrow\left\{{}\begin{matrix}x+y=-\left(z+t\right)\\y+z=-\left(x+t\right)\\z+t=-\left(x+y\right)\\t+x=-\left(y+z\right)\end{matrix}\right.\)

\(P=\dfrac{x+y}{z+t}+\dfrac{y+z}{t+x}+\dfrac{z+t}{x+y}+\dfrac{t+x}{y+z}\\ =\dfrac{-\left(z+t\right)}{z+t}+\dfrac{-\left(t+x\right)}{t+x}+\dfrac{-\left(x+y\right)}{x+y}+\dfrac{-\left(y+z\right)}{y+z}\\ =-1-1-1-1\\ =-4\)

Tham khảo: https://hoc24.vn/cau-hoi/cho-bieu-thuc-pdfracxyztdfracyztxdfracztxydfractxyz-tinh-gia-tri-bieu-thuc-p-biet-dfracxyztdfracyzt.3023321885549