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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?

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  • $\begingroup$ What's the quotient $n^3/(3n^2)$? $\endgroup$ – Lord Shark the Unknown Apr 15 '17 at 7:07
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    $\begingroup$ 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. $\endgroup$ – Olivier Oloa Apr 15 '17 at 7:10
  • $\begingroup$ How do you know f (x)>=0 for x >=0? That seems circular to me. $\endgroup$ – fleablood Apr 15 '17 at 7:29
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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 $n^3\geq 3n^2$.

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  • $\begingroup$ I was just going to post an answer with exactly the same thing as you wrote but you beat me to it :) $\endgroup$ – user214302 Apr 15 '17 at 7:11
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For $n\geq3$, note $n^3=n\cdot n^2\geq3n^2$

Induction is a bit overkill for this question.

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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 $n=0$.)

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