Understanding a proof from the APMO 1998 on inequalities.

Question

I was having trouble with proving the following inequality.The question was from the book Secrets to Inequalities by Pham Kim Hung.

$\frac{x}{y} + \frac{y}{z} + \frac{z}{x} \geq \frac{x+y+z}{\sqrt[3]{xyz} }$

I know that the AM-GM inequality must be applied but I am having a hard time proving the the above inequality. In the book the author was able to show that by applying the trick shown below and then applying AM-GM we can prove the inequality holds.

$3\left (\frac{x}{y}+\frac{y}{z}+\frac{z}{x} \right ) = \left ( \frac{2x}{y} +\frac{y}{z}\right ) + \left ( \frac{2y}{z}+\frac{z}{x} \right ) + \left ( \frac{2z}{x}+\frac{x}{y} \right ) \geq \frac{3x}{\sqrt[3]{xyz}} + \frac{3y}{\sqrt[3]{xyz}} + \frac{3z}{\sqrt[3]{xyz}}$

The problem I had was after applying the AM-GM to the LHS I could not prove the RHS.

Answer 1

It is by AM-GM: $$\frac{x}{y}+\frac{x}{y}+\frac{y}{z}\geq 3\sqrt[3]{\frac{x^2y}{y^2z}}=3\sqrt[3]{\frac{x^3}{xyz}}$$

Answer 2

You can get the Pham Kim Hung's proof by the following way.

We'll try to find non-negatives $a$, $b$ and $c$ such that $a+b+c=1$ and $$a\cdot\frac{x}{y}+b\cdot\frac{y}{z}+c\cdot\frac{z}{x}\geq\frac{x}{\sqrt[3]{xyz}}$$ Now, by AM-GM $$a\cdot\frac{x}{y}+b\cdot\frac{y}{z}+c\cdot\frac{z}{x}\geq\left(\frac{x}{y}\right)^a\left(\frac{y}{z}\right)^b\left(\frac{z}{x}\right)^c=x^{a-c}y^{b-a}z^{c-b}.$$ Thus, we need $$x^{a-c}y^{b-a}z^{c-b}=\frac{x}{\sqrt[3]{xyz}}$$ or $$x^{a-c}y^{b-a}z^{c-b}=x^{\frac{2}{3}}y^{-\frac{1}{3}}z^{-\frac{1}{3}},$$ which gives the following system. $$a-c=\frac{2}{3},$$ $$b-a=-\frac{1}{3}$$ and $$a+b+c=1.$$ After solving of this system we obtain: $$(a,b,c)=\left(\frac{2}{3},\frac{1}{3},0\right),$$ which gives that by AM-GM $$\frac{2x}{3y}+\frac{y}{3z}\geq\frac{x}{\sqrt[3]{xyz}}$$ and since $$\frac{x}{y}+\frac{y}{z}+\frac{z}{x}=\sum_{cyc}\frac{x}{y}=\sum_{cyc}\left(\frac{2x}{3y}+\frac{y}{3z}\right)\geq\sum_{cyc}\frac{x}{\sqrt[3]{xyz}}=\frac{x+y+z}{\sqrt[3]{xyz}},$$ we are done!