For which values of $n$ does the following inequality stand? For which values of $n$, does it stand that:
$$-3n^4\log^2 n \geq -3 n^4 \sqrt{n} \quad ?$$
 A: Doing the easy cancellations for the $-3n^4$ factors you get the equivalent statement

$$\log^2n\le \sqrt n$$

i.e. $4\log\log n\le \log n$
setting $x=\log n$ we have $4\log x\le x$ i.e. ${x\over\log x}\ge 4$. But this function is always increasing  since the derivative of ${x\over\log x}$ is just ${\log x-1\over\log^2 x}>0$ for every $x>e$, i.e. every natural number, $n$, greater than $e^e\approx 15$, so just testing some values gives $x\approx 8.6135$ so then $e^{8.6135}\approx 5505\approx n$.
If you're looking for the smallest, there's some poking around to be done, since this is not exact, and the value is gotten by testing some values for ${x\over\log x}$, but doing it for $x=8.6135$ gets accuracy to three decimal places (for $x$), and poking around, one sees that $5504$ works and $5503$ does not, so $5504$ is the smallest.

As Daniel notes, the above discussion is when it is true forevermore, i.e. after that point we're going to see this inequality hold. For specific values less than $e^e$, i.e. $n\le 15$ we can just test directly and it turns out the inequality holds for $n\in \{1,2, 3, 4\}$ as well.
