If $x>0$, then find the greatest value of the expression $ \frac{x^{100}}{1+x+x^2+x^3+\cdots+x^{200}}$ 
If $x>0$, then find the greatest value of the expression $ \dfrac{x^{100}}{1+x+x^2+x^3+\cdots+x^{200}}$     

This expression simplifies to $ \frac{(x^{100})(x-1)}{x^{200}-1}$ using sum of n terms of GP. Now one can find the maxima by equating the derivative to zero. But is there any other way not involving calculus to get the maximum value?
 A: Hint: write it as
$$
\begin{align}
& \quad \frac{1}{\cfrac{1}{x^{100}} + \cfrac{1}{x^{99}} + \cdots + \cfrac{1}{x} + 1 + x + \cdots + x^{99} + x^{100}} \\
&= \frac{1}{1 + \left(x+\cfrac{1}{x}\right) + \left(x^2+\cfrac{1}{x^2}\right) + \cdots +  \left(x^{100}+\cfrac{1}{x^{100}}\right)}
\end{align}
$$
and use the fact that $a+\frac{1}{a} \ge 2$ for $a \gt 0$, with equality iff $a=1$.
A: Using AM-GM at the denominator $\geq 201 ( {\displaystyle \prod_{i =0}^{200} x^i} )^{1/201} $ thus you get denominator is always more than $201 \cdot x^{100}$ thus the max value is $\frac{1}{201} $ at $1$ you can also use graphs to see that.
A: AM-GM inquelity $=>$
$$
\\1+x+\cdots+x^{2n}\geq (2n+1)(x^{0+1+\cdots+2n})^{1/{2n}}=(2n+1){x^{\frac{2n+1}{2}}}
\\f(x)=\frac{x^n}{1+x+\cdots+x^{2n}}\leq\frac{x^n}{(2n+1){x^{\frac{2n+1}{2}}}}=\frac{1}{(2n+1)\sqrt{x}}
\\
\\\max_{x>0}(f(x))=\frac{1}{(2n+1)\sqrt{x}}
$$
then $f(x)=\frac{1}{(2n+1)\sqrt{x}}\>$ if and only if $x=1$
$$\\$$
ANSWER $\>\frac{1}{2n+1}=\frac{1}{201}\>$.
(Sorry english my second language)
