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Find all natural numbers n so that

$$ n^3 +(n+1)^3 >(n+2)^3$$

I have to do induction to prove this. I expanded and simplified both sides, am I on the right track? I don't know what to do from here

user209663
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user677216
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    What did you obtain when you expanded and simplified? Can you do any part of the induction? Unfortunately, it sounds like you’re asking others to do your homework. Give it a shot and tell us where you got. – Steve Kass Jun 13 '19 at 00:56
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    I got 2n^3 + 3n^2 + 3n+1 > n^3 + 6n^2 + 12n +8. however I tried proving the base case and its false for n = 1 – user677216 Jun 13 '19 at 00:57
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    So the statement is not true for $n=1$ as you observed. It should be for $n > 5$ – user209663 Jun 13 '19 at 01:10
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    The inductive argument doesn't care where you start as long as you start in a valid place. Sounds like $n > 5$ is it. Now you just do your induction. Assume it's true for values of $n$ from 6 through $k$ then prove it's true for $k+1$. Plug in the expressions and work out the algebra. – lurker Jun 13 '19 at 01:24
  • Note: $2n^3+3n^2+3n+1=n^3+6n^2+12n+6$ when $n=5$ and $2n^3+3n^2+3n+1=n^3+6n^2+12n+55$ when $n=6$ – J. W. Tanner Jun 13 '19 at 02:21

4 Answers4

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Another hint...

Sounds like $n > 5$ is your starting point, but that doesn't affect the inductive process. You just assume it's true for values of $n$ from $6$ through $k$, then prove it's true for $k+1$ using that assumption. In other words, you assume $k^3 + (k+1)^2 > (k+2)^2$ and use that to show that $(k+1)^3 + (k+2)^3 > (k+3)^3$.

Looks like, from your comment, you plugged in $k+1$ to both sides of the inequality, but that's what you're trying to prove, not what you want to assume. Try just starting with the left side: $(k+1)^3+(k+2)^3$ then manipulate it to get to a form where you can use what you know is true in your inductive assumption:

$(k+1)^3+(k+2)^3 = k^3 + (k+1)^3 + (k+2)^3 - k^3$

Notice we purposely obtained a form we know something about: $k^3 + (k+1)^3$. Try going from there to show the above is $> (k+3)^3$. The fact that 6 is the smallest value you're assuming for $k$ may come in handy.

lurker
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From $2n^3 + 3n^2 + 3n+1 > n^3 + 6n^2 + 12n +8 <=> n^3 - 3n^2 - 9n > 7 <=> n(n^2 - 3n - 9) > 7$. the positive root of the polynomial in the brackets is $\frac{3}{2}(1+\sqrt{5}) \approx 4.85$. The polynomial is increasing to the right of this root => $(n+1)((n+1)^2 - 3(n+1) - 9) > n(n^2 - 3n - 9)$ for $n \geq 5$.

The induction is: check n(n^2 - 3n - 9) > 7 for n = 6

assume n(n^2 - 3n - 9) > 7 for some n > 6

=> $(n+1)((n+1)^2 - 3(n+1) - 9) > n(n^2 - 3n - 9) > 7.$

Grada Gukovic
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Hint:

enter image description here

As you commented that your base case doesn't work... and here is the reason why. So start with a different $n$ value (can you see what is the base case should be from the graph?) and see where you can go from there!

user209663
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Let there is a equation $x^3 + (x + 1)^3 - (x + 2)^3 = 0$ or $x^3 - 3x^2 - 9x - 7 = 0$. Its real root is $x = 1 + \sqrt[3]{9 - \sqrt{17}} + \sqrt[3]{9 + \sqrt{17}} > 5$. Thereby $n^3 + (n + 1)^3 > (n+2)^3$ if the $n \ge \lceil x \rceil = 6$.

David Taylor
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