<s>[INST] How to apply the AM-GM inequality? What is the minimum value of $8x^3+36x+54/x+27/x^3 $ for positive real numbers x?
Express your answer in simplest radical form.
I attempted to make an equation between the product of the terms and the original expression, $ 8*36*54*27 = 8x^3+36x+54/x+27/x^3 $ 
but it seems more complicated than necessary. Is there a better way to apply the inequality?
 [/INST] For numbers $a_1,\ldots,a_n > 0$, the AM-GM inequality is $\dfrac{a_1+a_2+\cdots+a_n}{n} \ge \sqrt[n]{a_1a_2\cdots a_n}$. 
Equality occurs iff $a_1 = a_2 = \cdots = a_n$. 
The trick here is to apply AM-GM to the right number of terms at a time. 
Since $x$ is positive, the $4$ terms $8x^3$, $36x$, $\dfrac{54}{x}$, and $\dfrac{27}{x^3}$ are all positive. 
Hence, $\dfrac{8x^3+36x+\tfrac{54}{x}+\tfrac{27}{x^3}}{4} \ge \sqrt[4]{8x^3 \cdot 36x \cdot \tfrac{54}{x} \cdot \tfrac{27}{x^3}}$
For equality to occur, you need $8x^3 = 36x = \dfrac{54}{x} = \dfrac{27}{x^3}$. Unfortunately, there is no such $x$ for which all four are equal. 
Instead, let's apply AM-GM to two terms at a time: 
$\dfrac{8x^3+\tfrac{27}{x^3}}{2} \ge \sqrt{8x^3 \cdot \tfrac{27}{x^3}} = \sqrt{8 \cdot 27} = 6\sqrt{6} \leadsto 8x^3+\tfrac{27}{x^3} \ge 12\sqrt{6}$
$\dfrac{36x+\tfrac{54}{x}}{2} \ge \sqrt{36x \cdot \tfrac{54}{x}} = \sqrt{36 \cdot 54} = 18\sqrt{6} \leadsto 36x+\tfrac{54}{x} \ge 36\sqrt{6}$
Now, add the two together to get $8x^3+36x+\tfrac{54}{x}+\tfrac{27}{x^3} \ge 48\sqrt{6}$. 
Finally, equality occurs iff $8x^3 = \tfrac{27}{x^3}$ and $36x = \tfrac{54}{x}$. There is in fact a positive number $x$ for which both are satisfied (specifically $x = \sqrt{\tfrac{3}{2}}$). 
Alternatively, if you notice that $8x^3+36x+\tfrac{54}{x}+\tfrac{27}{x^3} = \left(2x+\tfrac{3}{x}\right)^3$ as Alexey Burdin pointed out in the comments, then you can just apply AM-GM to $2x+\tfrac{3}{x}$. 
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