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:
\(VT\ge\frac{3}{x-1}-6+\frac{3x-3}{y}+\frac{3}{y}\)
\(=\frac{3-6x+6}{x-1}+\frac{3x}{y}\)
\(=3\left(\frac{3-2x}{x-1}+\frac{x}{y}\right)\)