\(Q=\frac{1}{\frac{x}{y}+\frac{z}{x}+1}+\frac{1}{\frac{y}{z}+\frac{x}{y}+1}+\frac{1}{\frac{z}{x}+\frac{y}{z}+1}\)
Đặt \(\left(\frac{x}{y};\frac{y}{z};\frac{z}{x}\right)=\left(a^3;b^3;c^3\right)\Rightarrow abc=1\)
\(Q=\frac{1}{a^3+c^3+1}+\frac{1}{a^3+b^3+1}+\frac{1}{b^3+c^3+1}\)
Ta có: \(a^3+b^3=\left(a+b\right)\left(a^2+b^2-ab\right)\ge\left(a+b\right)\left(2ab-ab\right)=ab\left(a+b\right)\)
\(\Rightarrow Q\le\frac{1}{ac\left(a+c\right)+1}+\frac{1}{ab\left(a+b\right)+1}+\frac{1}{bc\left(b+c\right)+1}\)
\(Q\le\frac{abc}{ac\left(a+c\right)+abc}+\frac{abc}{ab\left(a+b\right)+abc}+\frac{abc}{bc\left(b+c\right)+abc}\)
\(Q\le\frac{b}{a+b+c}+\frac{c}{a+b+c}+\frac{a}{a+b+c}=1\)
\(\Rightarrow Q_{max}=1\) khi \(a=b=c=1\) hay \(x=y=z\)