Áp dụng BĐT Cauchy, ta có:
\(VT\ge3\sqrt[3]{\dfrac{x^2.y^2.z^2}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}}=3\sqrt[3]{\dfrac{1}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}}\)
Ta có: xyz=1 và x,y,z >0
\(\Rightarrow x\le1\Rightarrow x+1\le2\Rightarrow\dfrac{1}{x+1}\ge\dfrac{1}{2}\)
Tương tự \(\dfrac{1}{y+1}\ge\dfrac{1}{2}\)
\(\dfrac{1}{z+1}\ge\dfrac{1}{2}\)
\(\Rightarrow VT\ge3\sqrt[3]{\dfrac{1}{x+1}.\dfrac{1}{y+1}.\dfrac{1}{z+1}}=\dfrac{3}{2}\)
Đẳng thức xảy ra khi x=y=z=1
Ta có \(x+y+z\ge3\sqrt[3]{xyz}=3.1=3\)
Theo BĐT Cauchy-Schwarz dạng Engel:
\(\dfrac{x^2}{y+1}+\dfrac{y^2}{z+1}+\dfrac{z^2}{x+1}\ge\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)+3}\)
\(\ge\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)+\left(x+y+z\right)}=\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\dfrac{x+y+z}{2}\ge\dfrac{3}{2}\)
Đẳng thức xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{y+1}=\dfrac{y}{z+1}=\dfrac{z}{x+1}\\xyz=1\end{matrix}\right.\)
\(\Leftrightarrow x=y=z=1\)
