How to find n in $n^3 > 3n^2 -100n +6$

Question

I tried by removing constant 6 to simplify my work:

$n^3>3n^2-100n$

$1> 3/n - 100/n^2 $

$n>3-100/n$

and then $ n + 100/n > 3$

I am stuck here any help, is there any standard way to find n from above equality.

Answer 1

Set $f(x) = x^3 - 3x^2 + 100x - 6$, then $f'(x) = 3x^2 - 6x + 100$.

Its not hard to show that $f'(x) > 0 \ \forall x$

Now the inequality isnt true for $n=0$ but it is true for $n=1$ and hence true $\forall n \in \mathbb{N}$ (Starting at $1$)

Answer 2

Here are a couple of manipulations which might help.

The first would be to make both sides positive, so to write the inequality as $$n^3+100n\gt 3n^2+6$$ now that becomes $$n(n^2+100)\gt 3n^2+6$$

Now this is obviously true whenever $n\ge 3$ because $3n^2+300\gt 3n^2+6$ and there are other cases to investigate.

Another would be to adjust to give a convenient factorisation on the right-hand side, something like $$n^3+91n\gt 3n^2-9n+6=3(n-1)(n-2)$$

Or you could make the left-hand side factor conveniently eg $$n^3+20n^2+100n\gt23n^2+6$$ with $$n(n+10)^2\gt 23n^2+6$$

I will leave you to explore the possibilities.