Olympiad Minimization Problem I've been struggling to find a solution to this problem that I found in the archive of my country's Olympiad questions. 
I'm particularly interested in a solution that doesn't involve the use of calculus since I know that Olympiad questions do not require the knowledge of calculus to solve but I will also like to see one that uses it.
Here's the problem:
Find the minimum value of $\frac{18}{a+b} + \frac{12}{ab} + 8a + 5b$  when $a$ and $b$ are positive real numbers.
 A: Use AM-GM by rearranging terms creatively
Hint: A good start of using AM-GM is to consider the following:   
$ \frac{ 12}{ab} + K a + L b \geq 3 \sqrt[3]{ 12 K L }$, with equality when  $ \frac{12}{ab} = K a = Lb$.
$ \frac{18}{a+b} + M(a+b) \geq 2 \sqrt{ 18 M }$, with equality when $\frac{18}{a+b} = M (a+b)$.
Now, pick suitable $K, L, M$, so that equality holds throughout for the same values of $a, b$.    
Hence, the minimum of the expression is ... which is achieved when ...

How to pick suitable $K, L, M$:
(I strongly encourage you to think about this before reading on. Write down whatever equations/motivations you can think of, 

 We want $ K + M  = 8, L + M = 5$.
 We wishfully think that $ 12 K L$ is a perfect cube, and $ 18 M$ is a perfect square.
 An obvious choice is $ M = 2, K = 6, L = 3$.
 We just need to verify that equality holds, and for the same values, which thankfully it does with $a = 1, b = 2$, giving the minimum value of 30.
 (Otherwise, do some other wishful thinking, pick some other value of $M$ and try again.)

A: For the calculus approach, setting the derivatives with respect to $a$ and $b$ to $0$ yields $$\frac{18}{(a + b)^2} + \frac{12}{a^2 b} = 8$$ and $$\frac{18}{(a + b)^2} 
 + \frac{12}{a b^2} = 5.$$
So 
$$\frac{12}{a^2 b} - \frac{12}{a b^2} = 3,$$
equivalently,
$$4(b - a) = a^2 b^2,$$
for which $(a,b)=(1,2)$ is an obvious solution that happens to satisfy both equations.
The second derivatives are all positive for positive $a$ and $b$.
