Proving or disproving inequality $ \frac{xy}{z} + \frac{yz}{x} + \frac{zx}{y} \ge x + y + z $ [duplicate]

Consider this inequality:

$$(a-b)^2 + (b-c)^2 + (c-a)^2 \ge 0$$ Expand and simplify the above expression to get:

$$a^2 + b^2 + c^2 \ge ab + bc + ca$$

Substitute $a = \dfrac{1}{x}, b = \dfrac{1}{y}$ and $c = \dfrac{1}{z}$:

$$\dfrac{1}{x^2} + \dfrac{1}{y^2} + \dfrac{1}{z^2} \ge \dfrac{1}{xy} + \dfrac{1}{yz} + \dfrac{1}{zx}$$

Multiply both sides with $xyz$ to get

$$\dfrac{xy}{z} + \dfrac{yz}{x} + \dfrac{zx}{y} \ge x + y + z$$

Use the rearrangement inequality:

$$xy={xyz\over z},$$ so WLOG(without loss of generality), if $0<x\le y\le z$, $$\begin{gather}\frac1x\ge\frac1y\ge\frac1z,\\ yz\ge zx\ge xy.\end{gather}$$ Then we have the result, since LHS of the given inequality is the maximum value when rearrange.

Holder's inequality:

$$u\cdot v \leq |u||v|$$

for any vectors $u,v$.

Let $\mathbf u=(1/x,1/y,1/z)$ and $\mathbf v=(xz,xy,yz)$. Then show $|v| = xyz |u|$ and thus $$|\mathbf u||\mathbf v| = xyz\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)= \dfrac{xy}{z} + \dfrac{yz}{x} + \dfrac{zx}{y} $$ and: $$\mathbf u\cdot\mathbf v = x+y+z $$