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

\(=\left(x+y+z\right)\left(xy+yz+zx\right)-\sqrt[3]{xyz}.\sqrt[3]{xy.yz.zx}\)

\(\ge\left(x+y+z\right)\left(xy+yz+zx\right)-\dfrac{1}{3}.\left(x+y+z\right).\dfrac{1}{3}\left(xy+yz+zx\right)\)

\(=\dfrac{8}{9}\left(x+y+z\right)\left(xy+yz+zx\right)\)

\(\ge\dfrac{8}{9}\sqrt{3\left(xy+yz+zx\right)}.\left(xy+yz+zx\right)\)

\(=\dfrac{8}{9}\sqrt{3\left(xy+yz+zx\right)^3}\)

\(\Rightarrow3\left(xy+yz+zx\right)^3\le\left(\dfrac{9}{8}\right)^2\)

\(\Rightarrow\left(xy+yz+zx\right)^3\le\dfrac{27}{64}\)

\(\Rightarrow xy+yz+zx\le\dfrac{3}{4}\)

\(\left(x^3+1\right)\left(y^3+1\right)\left(z^3+1\right)=\dfrac{81}{64}x^3y^3z^3\)

\(\Leftrightarrow\left(x+1\right)\left(y+1\right)\left(z+1\right)\left(x^2-x+1\right)\left(y^2-y+1\right)\left(z^2-z+1\right)=\dfrac{81}{64}x^2y^2z^2\)

\(\Leftrightarrow3xyz\left(x^2-x+1\right)\left(y^2-y+1\right)\left(z^2-z+1\right)=\dfrac{81}{64}x^3y^3z^3\)

 \(\Rightarrow\left[{}\begin{matrix}xyz=0\\\left(x^2-x+1\right)\left(y^2-y+1\right)\left(z^2-z+1\right)=\dfrac{27}{64}x^2y^2z^2\end{matrix}\right.\)

Nếu \(\left(x^2-x+1\right)\left(y^2-y+1\right)\left(z^2-z+1\right)=\dfrac{27}{64}x^2y^2z^2\) 

Ta có:

\(x^2-x+1=\dfrac{3}{4}x^2+\left(\dfrac{x}{2}-1\right)^2\ge\dfrac{3}{4}x^2\)

Tương tự: \(y^2-y+1\ge\dfrac{3}{4}y^2\) ; \(z^2-z+1\ge\dfrac{3}{4}z^2\)

Do các vế của các BĐT trên đều không âm, nhân vế với vế ta được:

\(\left(x^2-x+1\right)\left(y^2-y+1\right)\left(z^2-z+1\right)\ge\dfrac{27}{64}x^2y^2z^2\)

Đẳng thức xảy ra khi và chỉ khi \(x=y=z=\dfrac{1}{2}\) 

Thế vào  điều kiện \(\left(x+1\right)\left(y+1\right)\left(z+1\right)=3xyz\) ko thỏa mãn (loại)

Vậy \(xyz=0\)

Đặt a=x-2; b=y-2; c=z-2. Phải chứng minh abc =<1

Thật vậy, từ \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\)ta có:

\(\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}=1\)

Theo BĐT Cauchy ta có:

\(\frac{1}{a+2}=\left(\frac{1}{2}-\frac{1}{b+2}\right)+\left(\frac{1}{2}-\frac{1}{c+2}\right)=\frac{1}{2}\left(\frac{b}{b+2}+\frac{c}{c+2}\right)\ge\sqrt{\frac{bc}{\left(b+2\right)\left(c+2\right)}}\left(1\right)\)

tương tự \(\hept{\begin{cases}\frac{1}{b+2}\ge\sqrt{\frac{ac}{\left(a+2\right)\left(c+2\right)}}\left(2\right)\\\frac{1}{c+2}\ge\sqrt{\frac{ab}{\left(a+2\right)\left(b+2\right)}}\left(3\right)\end{cases}}\)

Nhân các vế của (1)(2)(3) ta được đpcm

Dấu "=" xảy ra <=> a=b=c hay x=y=z=3

Áp dụng BĐT BSC và BĐT Cosi:

\(17\left(x+y+z\right)+2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

\(\ge17\left(x+y+z\right)+\frac{2.\left(1+1+1\right)^2}{x+y+z}\)

\(=17\left(x+y+z\right)=\frac{18}{x+y+z}\)

\(=17\left(x+y+z\right)=\frac{17}{x+y+z}+\frac{1}{x+y+z}\)

\(\ge2\sqrt{17\left(x+y+z\right).\frac{17}{x+y+z}}+\frac{1}{1}\)

\(=35\)

\(\Rightarrow17\left(x+y+z\right)+2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge35\)

Đẳng thức xảy ra khi \(x=y=z=\frac{1}{3}\)

Áp dụng bất đẳng thức AM-GM kết hợp giả thiết x + y + z ≤ 1 ta có :

\(17\left(x+y+z\right)+2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=17x+17y+17z+\frac{2}{x}+\frac{2}{y}+\frac{2}{z}\)

\(=\left(18x+\frac{2}{x}\right)+\left(18y+\frac{2}{y}\right)+\left(18z+\frac{2}{z}\right)-\left(x+y+z\right)\)

\(\ge2\sqrt{18x\cdot\frac{2}{x}}+2\sqrt{18y\cdot\frac{2}{y}}+2\sqrt{18z\cdot\frac{2}{z}}-1=12\cdot3-1=35\)( đpcm )

Dấu "=" xảy ra <=> x=y=z=1/3