Prove that $ \frac{x}{yz} + \frac{y}{xz} + \frac{z}{xy} \geq \frac{1}{x} + \frac{1}{y} + \frac{1}{z} $ Prove that $ \frac{x}{yz} + \frac{y}{xz} + \frac{z}{xy}  \geq  \frac{1}{x} + \frac{1}{y} + \frac{1}{z} $ for $x,y,z \in \Bbb{R}$ and $xyz > 0 $.
I know that i can use the axioms of the real numbers, but i can't finde an usefull equivalent transformation 
 A: multiplying your given inequality by $$xyz>0$$ we get
$$x^2+y^2+z^2\geq xy+xz+zy$$
and this is equivalent to
$$(x-y)^2+(y-z)^2+(z-x)^2\geq  0$$ which is true.
A: Multiply through by $xyz$ and you get $$x^2 + y^2 + z^2 \geq xy + yz + zx.$$ Now note that $$(x-y)^2 + (y-z)^2 + (z-x)^2 \geq 0$$ with equality only if $x = y = z$.
A: By C-S
$$\sum_{cyc}\frac{x}{yz}=\sum_{cyc}\frac{x^2}{xyz}\geq\frac{(x+y+z)^2}{3xyz}\geq\frac{3(xy+xz+yz)}{3xyz}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$$
A: You can also use rearrangement inequality, which is true for all real numbers. The inequality is symmetric in terms of $x,y,z$. Let WLOG $x\ge y\ge z$. Notice that it's only given that $xyz>0$, $x,y,z\in\mathbb R$, not necessarily $x,y,z>0$. There are two cases:
$1)$ If $z>0$, then $y>0$, $x>0$.
$2)$ If $z<0$, then $y<0$, $x>0$.
In both cases $(x,y,z)$ and $\left(\frac{1}{yz},\frac{1}{xz},\frac{1}{xy}\right)$ are sorted in the same order.
$(x,y,z)$ and $\left(\frac{1}{xy},\frac{1}{yz},\frac{1}{zx}\right)$ is another way to sort them, but we don't care if they're sorted in the same or opposite or another order.
$$x\cdot\frac{1}{yz}+y\cdot\frac{1}{xz}+z\cdot \frac{1}{xy}\ge $$
$$\ge x\cdot\frac{1}{xy}+y\cdot\frac{1}{yz}+z\cdot \frac{1}{zx}=$$
$$=\frac{1}{y}+\frac{1}{z}+\frac{1}{x}$$
Edit: as other answers have noticed, multiplying both sides by $xyz>0$ gives $x^2+y^2+z^2\ge xy+yz+zx$ and there has been a proof shown. Another proof is using rearrangement inequality. $(x,y,z)$ and $(x,y,z)$ are sorted in the same order. $(x,y,z)$ and $(y,z,x)$ is another way to sort them.
A: since $x,y,z\in\mathbb R$
$$(x-y)^2+(y-z)^2+(z-x)^2\geq0\\
x^2+y^2+z^2\geq xy+yz+zx\\
\frac{x}{yz} + \frac{y}{xz} + \frac{z}{xy} \geq \frac{1}{x} + \frac{1}{y} + \frac{1}{z}$$ 
