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I know that it holds for $n=5$, so the first step is done.

For the second step, my IH is: $4^n > n^4$, and I must show that $4^{n+1} > (n+1)^4$. I did as follows:

$4^{n+1} = 4*4^n > 4n^4$, and if I show that $4n^4 > (n+1)^4$ holds for $n\geq 5$ I'm done, so I expand the right hand side and arrive to:

$$4n^4 > n^4 + 4n^3 + 6n^2 + 4n + 1$$

And this holds iff: $$3n^4 - 4n^3 - 6n^2 - 4n - 1 >0$$ And here I'm stuck, I don't really know how to show that the given inequality holds. Any hints?

Miguelgondu
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    For $n\geq 5$ it should. – Miguelgondu Sep 16 '14 at 19:49
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    If you really insist that you want to solve that problem by solving the inequality -- which may be the case if you are out of ideas -- then you my add to both sides of your inequality $4$ and you will get a classic example of a symmetric polynomial, you may get its roots (hence the possibility to study its signs over all of $\mathbb R$) by first checking that $n=0$ is not a solution, then dividing $3n^4 - 4n^3 - 6n^2 - 4n +3=0$ by $n^2$... (the rest of the procedure is detailed in Gelfand-Algebra p.117 which you may easily find freely online) But I'm not sure whether that approach will work... – Hakim Sep 16 '14 at 19:55
  • ^If you want me to write detailed answer about that general approach then just ask away, I'll do that. – Hakim Sep 16 '14 at 19:57
  • @Hakim Don't worry: I'll look it up by myself. Thanks. – Miguelgondu Sep 16 '14 at 19:58
  • @Miguelgondu You're welcome, keep us updated with your progress. – Hakim Sep 16 '14 at 20:11

4 Answers4

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We only need to show that $4\ge (\frac{n+1}{n})^4=(1+\frac{1}{n})^4$ for $n\ge 5$ to prove the induction step, and this only need be shown for $n=5$ because as $n$ increases, $(1+\frac{1}{n})^4$ decreases.

please delete me
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Well you are not far... $$ 3n^4-4n^3-6n^2-4n-1=(n^2-2n-1)(3n^2+2n+1) $$ and that is always greater than zero for $n\ge 3$ (easy to see). The second term is always greater than zero since n is positive, and the first one is always greater than zero for $n\ge 3$.

Make sense?

Umberto

Umberto
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  • Makes a lot of sense, thanks. Is there an easy way of getting to such factorization? – Miguelgondu Sep 16 '14 at 19:59
  • Well if you are interested you could check this http://math.stackexchange.com/questions/785/is-there-a-general-formula-for-solving-4th-degree-equations (but that is very complex). @Hakim gave a very nice way of finding it out. Or if you are lazy (ehmm) you could use wolfram alpha... – Umberto Sep 16 '14 at 20:03
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Another way to show that $3n^4>4n^3+6n^2+4n+1$ for $n\ge5$ is to use

$3n^4\ge15n^3=4n^3+11n^3\ge4n^3+55n^2=4n^3+6n^2+49n^2\ge4n^3+6n^2+245n$

$\;\;\;\;\;\ge4n^3+6n^2+4n+1$ for $n\ge5$ (although I prefer Will Hunting's approach).

user84413
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You could also split the inequality into 3 pieces:

$n^4 > 4n^3 \Longleftrightarrow n > 4$

$n^4 > 6n^2 \Longleftrightarrow n^2 > 6$

$n^4 > n+1 \Longleftrightarrow n^3 > 1/n$

You can easily see that the inequalities are all true, because $n$ is at least equal to $5$.

Daniel W. Farlow
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Imago
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