\(\left\{{}\begin{matrix}x+y+z=1\\\left(1+\dfrac{1}{x}\right)\left(1+\dfrac{1}{y}\right)\left(1+\dfrac{1}{z}\right)=64\end{matrix}\right.\)
Ta có:
\(1=x+y+z\ge3\sqrt[3]{xyz}\)
\(\Leftrightarrow xyz\le\dfrac{1}{27}\)
Ta có: \(\left(1+\dfrac{1}{x}\right)\left(1+\dfrac{1}{y}\right)\left(1+\dfrac{1}{z}\right)=1+\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}\right)+\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)+\dfrac{1}{xyz}\)
\(\ge1+\dfrac{3}{\sqrt[3]{x^2y^2z^2}}+\dfrac{3}{\sqrt[3]{xyz}}+\dfrac{1}{xyz}\)
\(=1+\dfrac{3}{\sqrt[3]{\dfrac{1}{27^2}}}+\dfrac{3}{\sqrt[3]{\dfrac{1}{27}}}+\dfrac{1}{\dfrac{1}{27}}=64\)
Dấu = xảy ra khi \(x=y=z=\dfrac{1}{3}\)
