"Find the minimum value of natural number n, for which the inequality : $\frac{a(1-a)}{n} < \frac{1}{10^3}$ holds true for all $ a \in \mathbb{R}$ ".
The first step seems to be rewriting the equation: $1000a (1 - a) < n.$ I then tried to get the inequality in a quadratic form and solve for Discriminant $ \geqq 0. $ but I seem to be getting something wrong, since the answer I got is $n \leqq 250. $ Which doesn't make much sense.
What am I doing wrong? Is there another, perhaps better, more intuitive approach to this problem?
We have $$ a(1-a)=\frac14-(a-\tfrac12)^2\le \frac14$$ with equality iff $a=\frac12$, so we want $n>250$, i.e., the smallest such $n$ is $n=251$.