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?

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)
