<s>[INST] If $x+y+z=6$ and $xyz=2$, then find $\cfrac{1}{xy} +\cfrac{1}{yz}+\cfrac{1}{zx}$ 
If $x+y+z=6$ and $xyz=2$, then find the value of $$\cfrac{1}{xy}
 +\cfrac{1}{yz}+\cfrac{1}{zx}$$

I've started by simply looking for a form which involves the given known quantities ,so:
$$\cfrac{1}{xy} +\cfrac{1}{yz} +\cfrac{1}{zx}=\cfrac{yz\cdot zx +xy \cdot zx +xy \cdot yz}{(xyz)^2}$$
Now this might look nice since I know the value of the denominator but if I continue to work on the numerator I get looped :
$$\cfrac{yz\cdot zx +xy \cdot zx +xy \cdot yz}{(xyz)^2}=\cfrac{4\left(\cfrac{1}{xy}+\cfrac{1}{zy}+\cfrac{1}{zy}\right)}{(xyz)^2}=\cfrac{4\left(\cfrac{(\cdots)}{(xyz)^2}\right)}{(xyz)^2}$$
How do I deal with such continuous fraction ?
 [/INST] $$x+y+z=6$$
Divide both sides with $xyz$
$$\frac{x+y+z}{xyz}=\frac{6}{xyz}$$
$$\frac{x}{xyz}+\frac{y}{xyz}+\frac{z}{xyz}=\frac{6}{2}$$
$$\frac{1}{yz}+\frac{1}{xz}+\frac{1}{xy}=3$$
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