\(\left(1+\dfrac{1}{x}\right)\left(1+\dfrac{1}{y}\right)\left(1+\dfrac{1}{z}\right)\ge64\)
\(\Leftrightarrow\dfrac{\left(x+1\right)\left(y+1\right)\left(z+1\right)}{xyz}\ge64\)
Do đó ta cần chứng minh : \(\left(x+1\right)\left(y+1\right)\left(z+1\right)\ge64xyz\)
Theo BĐT \(AM-GM\) ta có :
\(\left\{{}\begin{matrix}x+1=x+x+y+z\ge4\sqrt[4]{x^2yz}\\y+1=y+x+y+z\ge4\sqrt[4]{y^2zx}\\z+1=z+x+y+z\ge4\sqrt[4]{z^2xy}\end{matrix}\right.\)
Nhân vế theo vế ta được :
\(\left(x+1\right)\left(y+1\right)\left(z+1\right)\ge64xyz\)
\(-->đpcm\)
áp dụng BĐT cô si ta có :
ta có : \(\dfrac{yz}{x}+\dfrac{zx}{y}+\dfrac{zx}{y}+\dfrac{xy}{z}+\dfrac{xy}{z}+\dfrac{yz}{x}\ge2z+2x+2y\)
\(\Leftrightarrow2\dfrac{yz}{x}+2\dfrac{zx}{y}+2\dfrac{xy}{z}\ge2z+2x+2y\)
\(\Leftrightarrow\dfrac{yz}{x}+\dfrac{zx}{y}+\dfrac{xy}{z}\ge z+x+y\left(đpcm\right)\)
dấu "=" xảy ra khi \(\dfrac{yz}{x}=\dfrac{zx}{y}=\dfrac{xy}{z}\Leftrightarrow x=y=z\)