Question

Cho x,y,z >0. Tìm Min f(x,y,z) = \(\dfrac{\left(x+y+z\right)^{16}}{xy^2z^3}\)

 

Answer 1

Sửa đề: \(Minf\left(x,y,z\right)=\frac{\left(x+y+z\right)^6}{xy^2z^3}\)

\(\frac{\left(x+y+z\right)^6}{xy^2z^3}=\frac{\left(x+\frac{y}{2}+\frac{y}{2}+\frac{z}{3}+\frac{z}{3}+\frac{z}{3}\right)^6}{xy^2z^3}\)

\(\ge\frac{\left(6\sqrt[6]{x.\frac{y^2}{4}.\frac{z^3}{27}}\right)^6}{xy^2z^3}=\frac{6^6}{4.27}=432\)

Answer 2

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