<s>[INST] A surprising results with a interesting inequality let $x,y,z>0$ such $x+y+z=3$, Find the minimum of the value
$$F=\dfrac{1}{x}+\dfrac{1}{xy}+\dfrac{1}{xyz}$$
I use wolfapha found  the surprising results $F_{min}=\dfrac{3+\sqrt{5}}{2}$
see links
maybe use AM-GM inequality,but I can't it,so How to prove $F\ge\dfrac{3+\sqrt{5}}{2}$
The question came from a middle school exam (14 age ago)
 [/INST] If we allow for 1-variable calculus, then

*

*Fix $y+z = k$. Show that the minimum of $\frac{1}{y} + \frac{1}{yz} = \frac{z+1}{z(k-z) } $ subject to $ y> 0, z> 0$ has value


 $\frac{1}{ (\sqrt{k+1} -1 ) ^2 }$ and occurs at $z = \sqrt{k+1} - 1$.



*So $ \frac{1}{x} + \frac{1}{xy} + \frac{1}{xyz}  = \frac{1}{x} ( 1 + \frac{1}{y} + \frac{1}{yz} ) \geq \frac{1}{3-k} ( 1 + \text{ answer from above } ) $. We can verify that the minimum occurs at $ k = \frac{1 + \sqrt{5}}{2}$, and has value $\frac{ 3 + \sqrt{5} } { 2}$.

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