Doubt in solution of APMO 1998 Inequality problem Question -
Let $a, b, c$ be positive real numbers. Prove that
$$
\begin{array}{c}
\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right) \geq 2+\frac{2(x+y+z)}{\sqrt[3]{x y z}} \\
(\text { APMO } 1998)
\end{array}
$$
My doubt -
in pham kim hung secrets they proved like this -
Solution. Certainly, the problem follows the inequality
$$
\frac{x}{y}+\frac{y}{z}+\frac{z}{x} \geq \frac{x+y+z}{\sqrt[3]{x y z}}
$$
which is true by AM-GM because
$$
3\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\right)=\left(\frac{2 x}{y}+\frac{y}{z}\right)+\left(\frac{2 y}{z}+\frac{z}{x}\right)+\left(\frac{2 z}{x}+\frac{x}{y}\right) \geq \frac{3 x}{\sqrt[3]{x y z}}+\frac{3 y}{\sqrt[3]{x y z}}+\frac{3 z}{\sqrt[3]{x y z}}
$$
now i did not understand how they got to this
$\frac{x}{y}+\frac{y}{z}+\frac{z}{x} \geq \frac{x+y+z}{\sqrt[3]{x y z}}$
in starting not in end???
when i expand LHS i get total 6 reciprocal terms and 2 cancelled from both sides but i did not understand how they cancel other 2 on RHS and remaining 3 terms on LHS.......
thankyou

$$\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)
=2+\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\right)+\left(\frac{y}{x}+\frac{z}{y}+\frac{x}{z}\right).$$
And each term in parentheses satisfies the inequality you have proved.
