Find the maximum of $\frac{abc}{(4a+1)(9a+b)(4b+c)(9c+1)}$,where$a,b,c>0$ $a,b,c>0$, find the maximum of :
$$\frac{abc}{(4a+1)(9a+b)(4b+c)(9c+1)}$$

I try to find the minimum of $\frac{(4a+1)(9a+b)(4b+c)(9c+1)}{abc}=\frac{4a+1}{\sqrt{a}}\cdot\frac{9a+b}{\sqrt{ab}}\cdot\frac{4b+c}{\sqrt{bc}}\cdot\frac{9c+1}{\sqrt{c}}
=\left(4\sqrt{a}+\frac{1}{\sqrt{a}}\right)\left(9\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}\right)\left(4\sqrt{\frac{b}{c}}+\sqrt{\frac{c}{b}}\right)\left(9\sqrt{c}+\frac{1}{\sqrt{c}}\right) \geq 4 \times 6\times 4\times 6$
when I try to use AM-GM inequality, I find I can't take equality in the above $4$ parentheses.so I only find a upper bound of the expression.: $\frac{abc}{(4a+1)(9a+b)(4b+c)(9c+1)} < \frac{1}{576}$

Hint
Try instead to minimize
$$\Phi=\frac{(4a+1)(9a+b)(4b+c)(9c+1)}{abc}$$ Compute the partial derivatives
All of them being equal to $0$, the solution is immediate.
