What is the maximum value of $\frac{x^{100}}{1+x+x^2+\ldots+x^{200}}$?

The denominator (in the original form) is a multiple of an arithmetic mean - a sum of $201$ terms is $201$ times their average. So then, depending on taste, you can either apply AM-GM to the denominator or GM-HM to the whole thing.

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 $x=1$.

For $x>0$, we have from the AM-GM inequality

$$\begin{align} \sum_{n=0}^{200}x^n&\ge 201 \sqrt[201]{\prod_{n=0}^{200}x^n}\\\\ &=201 \sqrt[201]{x^{20100}}\\\\ &=201x^{100} \end{align}$$

Hence, we see that

$$\frac{x^{100}}{\sum_{n=0}^{200}x^n}\le \frac1{201}$$