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.: $\frac{abc}{(4a+1)(9a+b)(4b+c)(9c+1)} < \frac{1}{576}$
 A: 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.
A: Let $Q = \dfrac{abc}{(4a+1)(9a+b)(4b+c)(9c+1)}= \dfrac{abc}{(b+9a)(4a+1)(4b+c)(9c+1)}$. Apply Cauchy-Schwarz inequality twice: $Q \le \dfrac{abc}{(2\sqrt{a}\cdot \sqrt{b}+1\cdot 3\sqrt{a})^2\cdot(2\sqrt{b}\cdot 3\sqrt{c}+\sqrt{c}\cdot 1)^2} = \left(\dfrac{\sqrt{b}}{(2\sqrt{b}+3)(6\sqrt{b}+1)}\right)^2= \left(\dfrac{t}{(2t+3)(6t+1)}\right)^2$, with $t = \sqrt{b}$. And apply AM-GM inequality:$\left(\dfrac{t}{(2t+3)(6t+1)}\right)^2= \left(\dfrac{t}{12t^2+20t+3}\right)^2= \dfrac{1}{\left(12t+\dfrac{3}{t}+20\right)^2}\le \dfrac{1}{\left(20+2\sqrt{12t\cdot \dfrac{3}{t}}\right)^2}=\dfrac{1}{32^2}=\dfrac{1}{1024}$. Thus $Q_{\text{max}} = \dfrac{1}{1024}$. This maximum value occurs when: $12t = \dfrac{3}{t}\implies t = \dfrac{1}{2}\implies \sqrt{b} = \dfrac{1}{2}\implies b = \dfrac{1}{4}$. Also: $\dfrac{2\sqrt{b}}{3\sqrt{c}}=\dfrac{\sqrt{c}}{1}\implies 3c=2\sqrt{b}=2\cdot\dfrac{1}{2} = 1\implies c = \dfrac{1}{3}.$, and $\dfrac{2\sqrt{a}}{\sqrt{b}}=\dfrac{1}{3\sqrt{a}}\implies 6a=\sqrt{b}=\dfrac{1}{2}\implies a = \dfrac{1}{12}.$ In summary, the maximum of $\dfrac{1}{1024}$ achieved when $(a,b,c) = \left(\dfrac{1}{12}, \dfrac{1}{4}, \dfrac{1}{3}\right)$.
