TITLE: Find the maximum value of $\frac{xyz}{(1+5x)(4x+3y)(5y+6z)(z+18)}$
QUESTION [5 upvotes]: Find the maximum value of $\frac{xyz}{(1+5x)(4x+3y)(5y+6z)(z+18)}$ as $x$, $y$, and $z$ range over all positive real numbers.
My first instinct was to apply AM-GM on each factor in the denominator since every variable is a positive real number. I also thought that it is just right to do so because minimizing the denominator would maximize the expression. The answer I got was $\frac{1}{2880}$, but the answer key states that it is $\frac{1}{5120}$. How come?

REPLY [4 votes]: Just another way, using an extended version of Holder’s inequality (or CS inequality repeatedly):
$$(1+5x)(4x+3y)(5y+6z)(z+18) \geqslant \left((1\cdot4x\cdot5y\cdot z)^{1/4}+(5x\cdot3y\cdot6z\cdot18)^{1/4} \right)^4 = 5120xyz$$
As equality is possible when $(x, y, z)=(\frac35, \frac{12}5, 6)$, this means the desired maximum is $\dfrac1{5120}$.