Ta có : 

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)

\(\Leftrightarrow\)\(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^3=0^3\)

\(\Leftrightarrow\)\(\left(\frac{1}{x}\right)^3+\left(\frac{1}{y}\right)^3+\left(\frac{1}{z}\right)^3+3\left(\frac{1}{x}+\frac{1}{y}\right)\left(\frac{1}{y}+\frac{1}{z}\right)\left(\frac{1}{z}+\frac{1}{x}\right)=0\)

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

Lại có : 

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)

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

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

\(\Leftrightarrow\)\(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{3}{xyz}\) ( đpcm ) 

Vậy nếu \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\) thì \(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{3}{xyz}\)

Chúc bạn học tốt ~ 

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Leftrightarrow\frac{1}{x}+\frac{1}{y}=\frac{-1}{z}\)

\(\Rightarrow\left(\frac{1}{x}+\frac{1}{y}\right)^3=\left(-\frac{1}{z}\right)^3\Leftrightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{3}{x^2y}+\frac{3}{xy^2}=-\frac{1}{z^3}\)

\(\Leftrightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{-3}{x^2y}-\frac{3}{xy^2}=\frac{-3}{xy}.\left(\frac{1}{x}+\frac{1}{y}\right)=\frac{-3}{xy}.-\frac{1}{z}=\frac{3}{xyz}\)

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Rightarrow\frac{1}{x}+\frac{1}{y}=-\frac{1}{z};\frac{1}{x}+\frac{1}{z}=-\frac{1}{y};\frac{1}{y}+\frac{1}{z}=-\frac{1}{x}\)

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

\(=\left(\frac{1}{x}+\frac{1}{y}\right)\left(\frac{1}{x^2}-\frac{1}{xy}+\frac{1}{y^2}\right)+\left(\frac{1}{x}+\frac{1}{z}\right)\left(\frac{1}{x^2}-\frac{1}{xz}+\frac{1}{z^2}\right)+\left(\frac{1}{y}+\frac{1}{z}\right)\left(\frac{1}{y^2}-\frac{1}{yz}+\frac{1}{z^2}\right)\)

\(=-\frac{1}{z}\left(\frac{1}{x^2}-\frac{1}{xy}+\frac{1}{y^2}\right)-\frac{1}{y}\left(\frac{1}{x^2}-\frac{1}{xz}+\frac{1}{z^2}\right)-\frac{1}{x}\left(\frac{1}{y^2}-\frac{1}{yz}+\frac{1}{z^2}\right)\)

\(=-\frac{1}{x^2z}+\frac{1}{xyz}-\frac{1}{y^2z}-\frac{1}{x^2y}+\frac{1}{xyz}-\frac{1}{yz^2}-\frac{1}{xy^2}+\frac{1}{xyz}-\frac{1}{xz^2}\)

\(=\left(-\frac{1}{x^2z}-\frac{1}{x^2y}\right)+\left(-\frac{1}{xy^2}-\frac{1}{y^2z}\right)+\left(-\frac{1}{xz^2}-\frac{1}{yz^2}\right)+\frac{3}{xyz}\)

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

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

\(\Rightarrow2\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)=\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}+\frac{3}{xyz}\Rightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{3}{xyz}\)(đpcm)

BĐT\(\Leftrightarrow\left(\frac{1}{x-1}\right)^3+\left(\frac{x-1}{y}\right)^3+\left(\frac{1}{y}\right)^3\ge3\left(\frac{1}{x-1}+\frac{x-1}{y}+\frac{1}{y}-2\right)\)

Đặt \(\left(\frac{1}{x-1};\frac{x-1}{y};\frac{1}{y}\right)=\left(a;b;c\right)\)

BĐT cần cm \(\Leftrightarrow a^3+b^3+c^3\ge3\left(a+b+c-2\right)\)

\(\Leftrightarrow\left(a^3+1+1\right)+\left(b^3+1+1\right)+\left(c^3+1+1\right)\ge3\left(a+b+c\right)\)

Đúng theo AM-GM --> đpcm

Biến đổi \(\frac{x}{y^3-1}-\frac{y}{x^3-1}=\frac{x^4-x-y^4+y}{\left(y^3-1\right)\left(x^3-1\right)}=\frac{\left(x^4-y^4\right)-\left(x-y\right)}{xy\left(y^2+y+1\right)\left(x^2+x+1\right)}\)

(Do x+y=1 => \(\hept{\begin{cases}y-1=-x\\x-1=-y\end{cases}}\))

\(=\frac{\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)-\left(x-y\right)}{xy\left(x^2y^2+y^2x+y^2+yx^2+xy+y+x^2+x+1\right)}\)

\(=\frac{\left(x-y\right)\left(x^3+y^3-1\right)}{xy\left[x^2y^2+xy\left(x+y\right)+x^2+y^2+xy+2\right]}\)

\(=\frac{\left(x-y\right)\left(x^2-x+y^2-y\right)}{xy\left[x^2y^2+\left(x+y\right)^2+2\right]}=\frac{\left(x-y\right)\left[x\left(x-1\right)+y\left(y-1\right)\right]}{xy\left(x^2y^2+3\right)}\)

\(=\frac{\left(x-y\right)\left[x\left(-y\right)+y\left(-x\right)\right]}{xy\left(x^2y^2+3\right)}=\frac{\left(x-y\right)\left(-2xy\right)}{xy\left(x^2y^2+3\right)}=\frac{-2\left(x-y\right)}{x^2y^2+3}\)

\(\Rightarrow\frac{x}{y^3-1}-\frac{y}{x^3-1}+\frac{2\left(x-y\right)}{x^2y^2+3}=0\left(đpcm\right)\)

Áp dụng BĐT cô si\(\frac{1}{\left(x-1\right)^3}+1+1\ge\sqrt[3]{\frac{1}{\left(x-1\right)^3}\cdot1\cdot1}=\frac{1}{x-1}\)

\(\Rightarrow\frac{1}{\left(x-1\right)^3}\ge\frac{3}{x-1}-2\left(1\right)\)

\(\left(\frac{x-1}{y}\right)^3+1+1\ge3\sqrt[3]{\left(\frac{x-1}{y}\right)^3\cdot1\cdot1}=\frac{3x-3}{y}\)

\(\Rightarrow\left(\frac{x-1}{y}\right)^3\ge\frac{3x-3}{y}-2\left(2\right)\)

\(\frac{1}{y^3}+1+1\ge\sqrt[3]{\frac{1}{y^3}\cdot1\cdot1}=\frac{3}{y}\Rightarrow\frac{1}{y^3}=\frac{3}{y}-2\left(3\right)\)

Cộng vế theo vế của \(\left(1\right);\left(2\right);\left(3\right)\) ta có:

\(VT\ge\frac{3}{x-1}-6+\frac{3x-3}{y}+\frac{3}{y}\)

\(=\frac{3-6x+6}{x-1}+\frac{3x}{y}\)

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

Ta có:

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

\(\ge3\left(\dfrac{1}{x-1}+\dfrac{x-1}{y}+\dfrac{1}{y}\right)\)

\(\Rightarrow\dfrac{1}{\left(x-1\right)^3}+\left(\dfrac{x-1}{y}\right)^3+\dfrac{1}{y^3}\ge3\left(\dfrac{1}{x-1}+\dfrac{x-1}{y}+\dfrac{1}{y}-2\right)\)

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

ĐK: \(x;y;z\ne0\)

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

\(=\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=0\left(đpcm\right)\)