令\(\frac{x}{y}={{a}^{3}},\frac{y}{z}={{b}^{3}},\frac{z}{x}={{c}^{3}}\)，則\(abc=1\)
\(\begin{align}
  & \frac{1}{1+\frac{x}{y}+\frac{y}{z}}=\frac{1}{abc+{{a}^{3}}+{{b}^{3}}}\le \frac{1}{abc+{{a}^{2}}b+a{{b}^{2}}}=\frac{1}{ab\left( a+b+c \right)}=\frac{c}{a+b+c} \\
& \frac{1}{1+\frac{y}{z}+\frac{z}{x}}\le \frac{a}{a+b+c} \\
& \frac{1}{1+\frac{z}{x}+\frac{x}{y}}\le \frac{b}{a+b+c} \\
& \frac{1}{1+\frac{x}{y}+\frac{y}{z}}+\frac{1}{1+\frac{y}{z}+\frac{z}{x}}+\frac{1}{1+\frac{z}{x}+\frac{x}{y}}\le 1 \\
\end{align}\)

由\(\frac{x}{y}\times \frac{y}{z}\times \frac{z}{x}=1\)
把分母的1用abc替換，為了得到a+b+c及滿足不等關係，所以假設為立方數