<s>[INST] Find the maximum value of $\frac{xyz}{(1+5x)(4x+3y)(5y+6z)(z+18)}$ 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?
 [/INST] By AM-GM
$$\frac{xyz}{(1+5x)(4x+3y)(5y+6z)(z+18)}=$$
$$=\frac{xyz}{\left(1+3\cdot\frac{5x}{3}\right)(4x+3\cdot y)(5y+3\cdot2z)(z+3\cdot6)}\leq$$
$$\leq \frac{xyz}{4\sqrt[4]{1\cdot\left(\frac{5x}{3}\right)^3}\cdot4\sqrt[4]{4x\cdot y^3}\cdot4\sqrt[4]{5y\cdot(2z)^3}\cdot4\sqrt[4]{z\cdot6^3}}=\frac{1}{5120}.$$
The equality occurs for $$1=\frac{5x}{3},$$ $$4x=y,$$ $$5y=2z$$ and $$z=6$$ or $$(x,y,z)=\left(\frac{3}{5},\frac{12}{5},6\right),$$ which says that we got a maximal value.
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