Question

Cho \(x,y,z>0\). CMR : \(\dfrac{\left(x+y+z\right)^6}{xy^2z^3}\ge\dfrac{1}{432}\)

Answer 1

Sửa đề: CMR: \(\dfrac{\left(x+y+z\right)^6}{xy^2z^3}\ge432\)

Ta có

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

\(\ge\dfrac{\left(6\sqrt[6]{x.\dfrac{y}{2}.\dfrac{y}{2}.\dfrac{z}{3}.\dfrac{z}{3}.\dfrac{z}{3}}\right)^6}{xy^2z^3}=\dfrac{6^6.\dfrac{xy^2z^3}{2^2.3^3}}{xy^2z^3}=\dfrac{6^6}{2^2.3^3}=432\)

Đẳng thức xảy ra \(\Leftrightarrow x=\dfrac{y}{2}=\dfrac{z}{3}\)