Minimum value of $\frac{(x+8)^3}{x}$ without calculus, and weighted AM-GM inequality So I was trying to find the minimum value of $f(x) = \frac{(x+8)^3}{x}, x>0$ without calculus but something weird happended when applying weighted AM-GM inequality. 
First of all, of course $f(x) = x^2 + 24x + 512 x^{-1} + 192$. So I decided to find the minimum by applying the weighted AM-GM inequality: $$1(x^2) + 1(24x) + 3 (\frac{512}{3} x^{-1}) + 1(192) \geq 6 (x^2 \times 24 x \times 512^3 3^{-3} x^{-3} \times 192)^{1/6} = 128 × 3^{5/6}.$$ Which is about 319.25. To my surprise, however, this is not the minimum value of the function (the minimum is actually f(x) = 432). Why did this happen? I usually do find correct values to the function minima when I apply the AM-GM inequality. What's different this time? 

You don't get the minimum because equality can not hold in your application of the AM-GM inequality:
$$
x^2 = 24 x = \frac{512}{3} x^{-1} = 192
$$
is not possible.
If you start with
$$
x+8 = x + 4 + 4 \ge 3 \sqrt[3]{x \cdot 4 \cdot 4}
$$
then you'll get
$$
 \frac{(x+8)^3}{x} \ge 27 \cdot 16 = 432
$$
which is sharp because equality holds for $x=4$.
