Proving or disproving inequality $ \frac{xy}{z} + \frac{yz}{x} + \frac{zx}{y} \ge x + y + z $ 
Given that $ x, y, z \in \mathbb{R}^{+}$, prove or disprove the
  inequality
$$ \dfrac{xy}{z} + \dfrac{yz}{x} + \dfrac{zx}{y} \ge x + y + z $$

I have rearranged the above to:
$$ x^2y(y - z) + y^2z(z - x) + z^2x(x - y) \ge 0 \\
\text{and, } \dfrac{1}{x^2} + \dfrac{1}{y^2} + \dfrac{1}{z^2} \ge \dfrac{1}{xy} + \dfrac{1}{xz} + \dfrac{1}{yz} $$
What now? I thought of making use of the arithmetic and geometric mean properties:
$$
\dfrac{x^2 + y^2 + z^2}{3} \ge \sqrt[3]{(xyz)^2} \\
\text{and, } \dfrac{x + y + z}{3} \ge \sqrt[3]{xyz}
$$
but I am not sure how, or whether that'd help me at all.
 A: 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
$$\dfrac{xy}{z} + \dfrac{yz}{x} + \dfrac{zx}{y} \ge x + y + z$$
A: 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:
$$\mathbf u\cdot\mathbf  v =  x+y+z $$ 
A: 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.
