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:

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

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

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

dùng BCS rồi đánh giá

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

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

\(\le3-\frac{9}{x+y+z+3}=3-\frac{9}{1+3}=\frac{3}{4}^{\left(đpcm\right)}\) (Áp dụng BĐT \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\))

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x+y+z=1\\\frac{1}{x+1}=\frac{1}{y+1}=\frac{1}{z+1}\end{cases}}\Leftrightarrow\hept{\begin{cases}x+y+z=1\\x+1=y+1=z+1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x+y+z=1\\x=y=z\end{cases}}\Leftrightarrow x=y=z=\frac{1}{3}\)

Chứng minh BĐT: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\) (a,b,c > 0)

Thật vậy,theo BĐT AM-GM (Cô si) ta có: \(VT\ge3\sqrt[3]{\frac{1}{abc}}=\frac{3}{\sqrt[3]{abc}}\ge\frac{3}{\frac{a+b+c}{3}}=\frac{9}{a+b+c}^{\left(đpcm\right)}\)

Day la duong thang Simson ma , con cau b dung duong thang Steiner la ra