Find the maximum of $\frac {xyz}{(1+x)(x+y)(y+z)(z+16)}$, given x, y, z are positive real numbers.

Question

Below is my working:
This is equivalent to the minimum of $$\frac {(1+x)(x+y)(y+z)(z+16)}{xyz}$$ $$=(1+x)\left(1+\frac{y}{x}\right)\left(1+\frac zy \right)\left( 1 + \frac {16}{z} \right)\\$$ $$= 1+ 16 + \frac {16}{z} + \frac zy + \frac yx + \frac {16y}{xz} +\frac zx + \frac {16y}{x} +x + \frac {16x}{z}+ \frac {xz}{y}+ 16x + y + \frac {16y}{z} + z + 16y \tag{1}\\ \text{By AM-GM}:\qquad \geq16\sqrt[16]{16^8} = 64$$
Thus the maximum value is $\frac {1}{64}$.
However, the solution gives $\frac {1}{81} $ as they used the generalization below : $$ (1+a^4)(1+b^4)(1+c^4)(1+d^4)\geq (1+abcd)^4 $$ Why is my answer incorrect? And how do I know when I can use AM-GM to find the minimum of an expression and when not to, if I don't know the generalization ?


EDIT: Now that I know the equality case never holds, how should I proceed with finding the minimum from (1)?

Answer 1

Your solution is incorrect because in order to achieve $1/64$, looking at the equality case, you should have $$1=16=\frac{16}{z}=\frac{z}{y}=\cdots$$which is obviously impossible. In general, if you prove that some expression satisfies $f(x,y,z)\geqslant\lambda$ you should also look if $\lambda$ can actually be achieved, i.e. your bounding wasn't too loose.

Let me give you an example (a random inequality I made up):

Let $a,b,c$ be positive real numbers such that $abc=1$. Prove that $$\frac{1}{2a+b+c}+\frac{1}{a+2b+c}+\frac{1}{a+b+2c}\leqslant \frac{3}{8}+\frac{a+b+c}{8}.$$

Intuitively, we see that the equality case is $a=b=c=1$. Hence, if we apply am-gm like this: $$2a+b+c\geqslant3\sqrt[3]{2abc}$$then we would have the equality case $2a=b=c$ so this inequality isn't sharp. Instead, we apply am-gm like this, which "preserves" the equality case (namely $a=b=c)$: $$2a+b+c\geqslant4\sqrt[4]{a^2bc}.$$Now, let's solve the inequality: \begin{align*}\sum_{\text{cyc}}\frac{1}{2a+b+c}&\leqslant\sum_{\text{cyc}}\frac{1}{4\sqrt[4]{a^2bc}}=\sum_{\text{cyc}}\frac{1}{4\sqrt[4]{a}}=\sum_{\text{cyc}}\frac{\sqrt[4]{bc}}{4} \\ &\leqslant\sum_{\text{cyc}}\frac{b+c+1+1}{16}=\frac{3}{8}+\frac{1}{8}\sum_{\text{cyc}}a.\end{align*}

Answer 2

You're literally almost there, but unfortunately you complicate the inequality too much. You have using AM-GM that $$ (1+x)^{-1}(1+y/x)^{-1}(1+z/y)^{-1}(1+16/z)^{-1}\leq\frac{1}{4}\left( (1+x)^{-4}+(1+y/x)^{-4}+(1+z/y)^{-4}+(1+16/z)^{-4}\right), $$ with equality iff $1+x = 1+y/x = 1+z/y = 1 + 16/z$. Now if you solve for $x,y,z$ and substitute back in the RHS of the inequality, you'll get the answer (it's messy).