How can I show without induction that $n^3\geq 3n^2$ for $n\geq3$?

Question

How can I show without induction that $n^3\geq 3n^2$ for $n\geq3$? ($n$ is a natural number).

My solution

$f(x)=x^3-3x^2\geq0$ for $x\geq 3$ so it follows that $n^3\geq 3n^2$ for $n\geq3$.

Any other solutions?

The inequality can be generalized as $f(x)\geq f'(x)$. Under what condition is this inequality true?

To adress your last question, the inequality $f'(x)\leq f(x)$ implies $f(x)\leq Ce^x$, for some constant $C$. If $f$ grows faster that $e^x$, then the inequality can't be true. — Olivier Oloa, Apr 15 '17 at 07:10
How do you know f (x)>=0 for x >=0? That seems circular to me. — fleablood, Apr 15 '17 at 07:29

Juniven Acapulco · Answer 1 · 2017-04-15T07:22:49.110

2

Because $n\geq 3$, then multiplying to both sides by $n^2$, we get $n\cdot n^2\geq 3\cdot n^2$, that is, $n^3\geq 3n^2$.

Alternately (as suggested in the comment of @Lord Shark ), if $n\geq 3$ then $\frac{n}{3}\geq 1$ and hence $$\frac{n^3}{3n^2}=\frac{n}{3}\geq 1$$ implying that $n^3\geq 3n^2$.

edited Apr 15 '17 at 07:22

answered Apr 15 '17 at 07:10

Juniven Acapulco

9,654

I was just going to post an answer with exactly the same thing as you wrote but you beat me to it :) – Apr 15 '17 at 07:11

score 1 · Answer 2 · answered Apr 15 '17 at 07:07

1

For $n\geq3$, note $n^3=n\cdot n^2\geq3n^2$

Induction is a bit overkill for this question.

answered Apr 15 '17 at 07:07

Mark Pineau

1,400

score 1 · Answer 3 · answered Apr 15 '17 at 07:08

1

For $n\neq 0$, $n^3\geq 3n^2$ if and only if $\frac{n^3}{n^2}\geq\frac{3n^2}{n^2}$, i.e. $n\geq 3$. This is because $n^2$ is positive, so you can divide/multiply both sides of an inequality by it. (The original inequality is also true for $n=0$.)

answered Apr 15 '17 at 07:08

Especially Lime

40,828

How can I show without induction that $n^3\geq 3n^2$ for $n\geq3$?

My solution

3 Answers3