How to find the optimal value of $x$ and what is the relation between $x$ and $n$ i.e. How to get dependency between $x$ and $n$?
As per my understanding, solution should be in term of $n$ like like $x= n/2$ or $2n$ or $>2n$ anything but it would be in term of $n$. I don't think so we will get any fixed answer.
I have solved it in one way: 1. Divide by $n^{2x}$ then 2. Let us assume: $n^{1-x}=y$ then got a quadratic equation: $y^2-y-1>0$. Further solved it and got $y>-\dfrac{1-\sqrt{5}}{2}$ and $y>\dfrac{1+\sqrt{5}}{2}$ i.e. y> -0.618 and y> 1.618. So, finally $n^{1-x}>1.618$.
How we can solve it in different way?
Let $y=x^{n-1}$. Then the given inequality can be written as: $y+y^2<1$, which holds for $a<y<b$, where $a,b$ are the roots of $t^2+t-1=0$. Since $y>0$, this is the same as: $x^{n-1}<b$, i.e. $x<b^{\frac{1}{n-1}}$ where $b=\dfrac{-1+\sqrt{5}}{2}$.