Find the minimum value of

$\large \frac{18}{a+b}+\frac{12}{ab}+8a+5b$

where
$a$
and
$b$
are
**
positive real numbers
**
.

The answer is 30.

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Rewrite the expression $\dfrac {18}{a+b} + \dfrac {12}{ab} + 8a+5b$ as follows and apply AM-GM inequality :

$\begin{aligned} {\color{#3D99F6}\frac {18}{a+b} + 2(a + b)} + {\color{#D61F06} \frac {12}{ab} + 6a+3b} \ge {\color{#3D99F6} 2\sqrt{\frac {18(2(a+b))}{a+b}}} + {\color{#D61F06} 3\sqrt[3]{\frac {12 (6a)(3b)}{ab}}} = {\color{#3D99F6}12} + {\color{#D61F06} 18} = \boxed{30} \end{aligned}$

Equality occurs when $a=1$ and $b=2$ .