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

Question

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?

Answer 1

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 $a=1$.

Answer 2

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.