Find all positive integers $a$ and $b$ such that $(1 + a)(8 + b)(a + b) = 27ab$. Here's the problem I'm having difficulties with:

Find all positive integers $a$ and $b$ such that $$(1 + a)(8 + b)(a + b) = 27ab\,.$$

Does anyone have an idea how to do this? Any detailed solution is welcome! :) 

Using Hölder's inequality,
$$27ab = (a+1)(8+b)(b+a) \geqslant \left(2\sqrt[3]{ab}+\sqrt[3]{ab} \right)^3=27ab$$
Hence we are looking for the equality case for Hölder, which is when $a:8:b=1:b:a \implies (a, b)=(2, 4)$.  
In fact, this is the only solution among positive reals, not just positive integers.
