Minimum of $x^2+8x+\frac{64}{x^3}$ for positive real $x$ 
Let $x$ be positive real number. Find the minimum value of $x^2+8x+\frac{64}{x^3}$.

I'm practicing some problems for a contest and this is one of them. Since most of the high school contests don't allow using calculus, I want to solve this without using calculus.
The problem looks obvious by AM-GM as $$x^2+8x+\frac{64}{x^3}\geq 3\sqrt[3]{8\cdot 64}=24$$
However, there is no $x$ for which $x^2=8x=\frac{64}{x^3}$ and thus this is not a minimum.
So how to deal with this problem?
 A: you can write the sum as $$ x^2 + 4 * 2x + 2* \dfrac{32}{x^3} $$
A: To use inequalities to find minima, we need the equality condition to be satisfied as well, hence in this case we need $x^2=8x=\frac{32}x$, which doesn't have a solution.
If you have a "lucky guess" for which $x$ the minimum occurs (and you know you have the minimum value somehow), we can split the terms to equal components and then apply AM-GM.  If not, there is still a way - consider the more general version:
$$x^2 + \alpha \; \frac{8x}\alpha + \beta\; \frac{64}{\beta x^3} \geqslant (1+\alpha+\beta) \left(x^2\cdot \left(\frac{8x}\alpha\right)^\alpha \cdot \left( \frac{64}{\beta x^3}\right)^\beta\right)^{\frac1{1+\alpha+\beta}} \tag{1}$$
While it is not necessary to write the above inequality out, I am reproducing it for clarity.
Now for using this inequality for minimum, the RHS must be a constant, i.e. $x-$free, so we need $\alpha = 3\beta-2$ considering the exponents of $x$.
Further, for the equality condition we must have some $x$ for which
$$x^2 = \frac{8x}\alpha = \frac{64}{\beta x^3} \iff x = \frac{8}{3\beta-2}  = \frac{64(3\beta-2)^4}{\beta \cdot 8^4}$$
The last equality gives $(3\beta-2)^5=2^9\beta$ which can be solved (well, by observation, or trying one's luck at rational roots) for $\beta=2$.  After doing all that, you can write out ($1$) with the correct $\alpha, \beta$ values for the desired AM-GM.
