TITLE: What is the maximum value of $\frac{x^{100}}{1+x+x^2+\ldots+x^{200}}$?
QUESTION [7 upvotes]: If $x$ is positive, what is the maximum value of this expression:
$$\frac{x^{100}}{1+x+x^2+\ldots+x^{200}}$$
This question is from a book of problems on sequence and series under the section on AM-GM-HM inequality.
This is what I have tried:
The denominator is a geometric series whose sum is
$$\frac{1-x^{201}}{1-x}$$
The fraction now becomes
$$\frac{x^{100}(1-x)}{1-x^{201}}$$
I can imagine that solving this problem will require taking the AM/GM/HM of some expressions of $x$ and applying the AM-GM-HM inequality.
That means the above fractions should themselves be one of GM or HM (whose maximum value will be given by the corresponding AM and GM respectively).
I can't see such means from looking at the fraction. Can someone help me here?

REPLY [5 votes]: You can instead minimise the reciprocal of your quantity, viz.,
$$\frac{1+x+x^2+\cdots+x^{200}}{x^{100}}=x^{-100}+x^{-99}+\cdots+x^{99}+x^{100}.$$
One only needs the two-variable AM/GM inequality to do this, just in the
form $y+y^{-1}\ge2$ for $y>0$, for
$$x^{-100}+x^{-99}+\cdots+x^{99}+x^{100}=1+\sum_{n=1}^{100}(x^n+x^{-n})
\ge201$$
with equality if $x=1$.