# Showing $2(n+5)^2 < n^3$

I stumbled upon this in my homework and can't seem to get it, any help would be great.

Find the smallest $n$ within $\mathbb N$ such that $2(n+5)^2 < n^3$ and call it $n_0$. Show that $2(n+5)^2 < n^3$ for all $n \geq n_0$.

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Can you use mathematical induction? – Doug Spoonwood Sep 6 '11 at 2:50
Yes I can use induction – Steve Sep 6 '11 at 2:53
Then you should use induction! – The Chaz 2.0 Sep 6 '11 at 3:05
The answers show you how to prove that once you have found $n_0$ the inequality is true for all higher $n$. To find $n_0$, you can just compute. A spreadsheet is ideal for this: a column for $2(n+5)^2$ and one for $n^3$, copy down, scan down, and there you are. If you are lazy, another column with the difference of the previous two. When it changes sign... – Ross Millikan Sep 6 '11 at 3:30

HINT $\$ Below are two sketched inductive proofs that $\rm\ f(n)\ =\ n^3- 2\ (n+5)^2\: >\: 0\$ for $\rm\ n \ge 7\:.$

Proof $\:1.$ $\rm\:\ \ f(n+1) - f(n)\ =\ (3\:n+8)\:(n-3)+3\: >\: 0\$ for $\rm\: n \ge 3\:,\:$ so $\rm\:f(n) < f(n+1)\:$ for $\rm\:n\ge 3\:.\:$ Therefore $\rm\ f(6) < 0 < f(7) < f(8) <\:\cdots\: < f(n)\$ for $\rm\:n \ge 7\:.$

Proof $\:2.$ $\rm\ \ f(n)>0\ \Leftrightarrow\ 1 < g(n) = \dfrac{n^3}{2\:(n+5)^2}\:.\:$ $\rm\ g(6) < 1 < g(7)\$ and $\rm\:g(n)\:$ is increasing by

$$\rm \frac{g(n+1)}{g(n)}\ =\ \frac{(n+1)^3}{n^3}\: \frac{(n+5)^2}{(n+6)^2}\ =\ \frac{n+1}{n}\ \bigg(\frac{n^2+6\ n+5}{n^2+6\ n}\bigg)^2\ >\ 1\quad for\quad n > 0$$

Essentially the $1$st proof is by additive telescopy, and the $2$nd by multiplicative telescopy. Notice how telescopy has reduced the induction to a trivial induction. Namely the first proof shows that $\rm\:f(n) > 0$ because it is a sum of terms $> 0\:,\:$ and the second proof shows that $\rm\:g(n) > 1$ because it is a product of terms $> 1\:.\:$ See the above linked posts for more on this viewpoint. The first additive telescopic method is essentially the fundamental theorem of difference calculus (whose proof - unlike the differential calculus form - is utterly trivial).

NOTE $\$ There are ad-hoc methods that are slightly faster than the above methods, e.g.

$$\rm n \ge \:8 \ \Rightarrow\ n^3 \ge \ 8\:n^2 =\: 2\:(n+n)^2 > \:2\:(n+5)^2$$

The point of mentioning the above telescopic techniques is that they have pedagogical value as general techniques for telescopic induction and, moreover, they lead to effective algortihms for much more general problems, e.g. computing closed forms for sums and products.

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precalculus tag probably indicates OP does not know how to take or use the derivative... – Ben Blum-Smith Sep 6 '11 at 3:11
@Ben The above proofs do not use derivatives or calculus. An earlier version included additionally a continuous form of the first proof, but since that has apparently led to some confusion, I have removed that one proof. – Bill Dubuque Sep 6 '11 at 4:43

Hint: $\frac{2(n+5)^2}{n^3} = \frac{2}{n} + \frac{20}{n^2} + \frac{50}{n^3}$ is a decreasing function of $n$ for $n > 0$.

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Rewrite $2(n+5)^2 < n^3$ as $2(1+\frac{5}{n})^2 < n$ , now you can see that for $n>5$ the left side is at most 8 so you know that the original form fail for $n=5$ and works for $n\geq8$. Test it for $n=6$ and $n=7$ and you are done.

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you can see that n=7 is the smallest natural number for which the statement is true.

Now try proving that $7n^2 \ge 2(n+5)^2$ for all $n \ge 7$ using the theory of quadratic expressions.

This is sufficient as $n^3 \ge 7n^2$ for $n \ge 7$

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