Find the maximum value of $\frac{xyz}{(1+5x)(4x+3y)(5y+6z)(z+18)}$

Question

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?

Answer 1

By AM-GM, we have $$1+5x \ge 2\sqrt{5x}, ~\text{ Eq. when}~ x=1/5~~~~(1)$$ $$4x+3y \ge 2\sqrt{12xy} ~\text{Eq. when}~y=4/15 ~~~~(2)$$ $$5x+6z \ge 2\sqrt{30xz} ~\text{Eq. when}~z=1/6 ~~~~~(3)$$ So $$z+18 > 2\sqrt{18z}~~~~ ~\text{ as Eq. when}~z=18 (contradiction)~~~~ (4)$$ Multiplying them we get $$(1+5x)(4x+3y)(5x+6z)(x+18) > 1440 xyz$$ $$\frac{xyz}{(1+5x)(4x+3y)(5x+6z)(x+18)} < \frac{1}{1440}.$$

Answer 2

It is equivalent to find the maximum for $w=\ln \frac{xyz}{(1+5x)(4x+3y)(5y+6z)(z+18)}$. Identify any critical point by setting $w_x'=w_y'=w_z'=0$, $$\frac1x=\frac5{1+5x}+\frac4{4x+3y}, \>\>\>\>\> \frac1y=\frac5{5y+6z}+\frac3{4x+3y}, \>\>\>\>\> \frac1z=\frac6{5y+6z}+\frac1{z+18}$$

Solve the equations for the point of positive solution at $(\frac35,\frac{12}5,6)$, which happens to admit the maximum value. Then, plug $(\frac35,\frac{12}5,6)$ into the expression to obtain

$$\frac{xyz}{(1+5x)(4x+3y)(5y+6z)(z+18)}=\frac1{5120}$$

Answer 3

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.

Answer 4

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 $\dfrac1{5120}$.