Find the minimum value of an expression with three variables

Question

How can I find the minimum of the following expression:

$(\frac{xy}{z}+\frac{zx}{y}+\frac{yz}{x})(\frac{x}{yz}+\frac{y}{xz}+\frac{z}{xy})$?

($x, y, z$ are non-zero real numbers)

The expression can be simplified to $\frac{((xy)^2+(zx)^2+(yz)^2)(x^2+y^2+z^2)}{(xyz)^2}$, but I am not sure that it will help anyhow.

Answer 1

By Cauchy-Schwarz inequality https://math.stackexchange.com/tags/cauchy-schwarz-inequality/info $$\left(\frac{xy}{z}+\frac{zx}{y}+\frac{yz}{x}\right)\left(\frac{x}{yz}+\frac{y}{xz}+\frac{z}{xy}\right)=\sum_{cyc}\frac{xy}{z}\sum_{cyc}\frac{z}{xy}\geq\left(\sum_{cyc}\sqrt{\frac{xy}{z}\cdot\frac{z}{xy}}\right)^2=9.$$ The equality occurs for $x=y=z$, which says that we gat a minimal value.

Also, we can use AM-GM: $$\sum_{cyc}\frac{xy}{z}\sum_{cyc}\frac{z}{xy}\geq3\sqrt[3]{\prod_{cyc}\frac{xy}{z}}\cdot3\sqrt[3]{\prod_{cyc}\frac{z}{xy}}=9.$$

Answer 2

Remember that for any 3 positive numbers we have $${a+b+c\over 3} \geq \sqrt[3]{abc}$$ this is inequality between arithmetic mean and geometic mean.

Use it twice. First $a= (xy)^2$, $b=....$

and second $a= x^2$,... and you get:

$$(\frac{xy}{z}+\frac{zx}{y}+\frac{yz}{x})(\frac{x}{yz}+\frac{y}{xz}+\frac{z}{xy})\geq 9$$

Answer 3

expanding gives $$3+\frac{x^2}{y^2}+\frac{y^2}{x^2}+\frac{x^2}{z^2}+\frac{z^2}{x^2}+\frac{y^2}{z^2}+\frac{z^2}{y^2}$$ and now use that $$A+\frac{1}{A}\geq 2$$ for $$A>0$$