Ta có : \(xyz=1\rightarrow\left\{{}\begin{matrix}xy=\dfrac{1}{z}\\xz=\dfrac{1}{y}\\yz=\dfrac{1}{x}\end{matrix}\right.\)
Do đó : \(A=\left(1+x\right)\left(1+y\right)\left(1+z\right)\)
\(A=1+x+y+z+xy+yz+xz+xyz\)
\(A=1+x+y+z+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+1\)
\(A=\left(x+\dfrac{1}{x}\right)+\left(y+\dfrac{1}{y}\right)+\left(z+\dfrac{1}{z}\right)+2\)
Áp dụng BĐT \(a+b\ge2\sqrt{ab}\left(a,b>0\right)\)
Dấu \(=\) xảy ra \(\Leftrightarrow a=b\)
với \(x,y,z>0\) Ta được :
\(A\ge2\sqrt{x.\dfrac{1}{x}}+2\sqrt{y.\dfrac{1}{y}}+2\sqrt{z.\dfrac{1}{z}}+2=2+2+2+2=8\)
Dấu \(=\) xảy ra \(\Leftrightarrow\)
\(\left\{{}\begin{matrix}x=\dfrac{1}{x}\\y=\dfrac{1}{y}\\z=\dfrac{1}{z}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2=1\\y^2=1\\z^2=1\end{matrix}\right.\Rightarrow x=y=z=1\) ( vì \(x,y,z>0\) )
