The answer isn't... yea

As of right now, I do not have a solution due to having found an error. I assumed that $xyz$ can be expressed in terms of $xy+yz+zx$ and $x+y+z$ , which isn't true. However, I haven't deleted the problem as I'm confident that the answer is correct through a graphing calculator (not a proof).

My approach was something like this: Assuming $xyz$ can be expressed in terms of $xy+yz+zx$ and $x+y+z$ , show that the expression is equivalent to:

$k\left( \left(\frac{x}{x+y+z}\right)^N + \left(\frac{y}{x+y+z}\right)^N+ \left(\frac{z}{x+y+z}\right)^N\right)$

Where $k$ is some real number. Then follow on with a generalised version of Khang's solution in this problem .

As for calculating the maximum value, I used $(x,y,z)=(1,1/4,1/4)$ , making the expression equal to:

$\frac{\frac{2}{2^{n(n+1)}}+1}{\prod _{k=1}^n\left(\frac{2}{4^k}+1\right)}$

Since $n=63$ is a pretty big number, the expression is very close to $\prod _{k=1}^n\left(\frac{2}{4^k}+1\right)^{-1}$ and it converges pretty fast. So it is easily calculatable on a calculator.

I haven't tried it but it seems like Alan's solution from the same problem can be generalised, albeit tediously

I can't do it because the compution is too much, but I know the exactly form of answer. Yeah, the f(t)=f(1/t) can't help anything It is mostly like although a term F(x,y) have a symmetry of x,y but can't say it reach extreme value at x=y. But for this question, it works , luckily.