Is there a solution for $(\frac{3}{x+y+z})^n+(\frac{3}{x+y+z})^{5-n}<2$? Is there a solution for
$$\left(\frac{3}{x+y+z}\right)^n+\left(\frac{3}{x+y+z}\right)^{5-n}<2$$
where $n\in\mathbb Z$ and $x,y,z>0, xyz=1$.
This is the part of my attempts for my homework, that I stucked.
This is obvious $x+y+z≥3\implies \frac {3}{x+y+z}≤1$. But there is no any restriction for integer $n$. Therefore, I stucked. I tried Wolfram Alpha, even WA couldn't solve.

Hint
By AM-GM inequality:
$$x+y+y\ge3(xyz)^{1/3}\Rightarrow \frac{3}{x+y+z}\le1.$$
The equality holds only when $x=y=z=1$.
It means that, for $0\le n \le 5$, we have
$$\left(\frac{3}{x+y+z}\right)^n+\left(\frac{3}{x+y+z}\right)^{5-n} \le 2.$$
You want to know if there is a solution for
$$\left(\frac{3}{x+y+z}\right)^n+\left(\frac{3}{x+y+z}\right)^{5-n} <2.$$
Think about the situation where you reach the sum equal to $2$.
Think on the interval $0\le n \le 5$.
Can you finish from here?
