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

Question

Here is the question: for which natural numbers do the following inequalities hold true? State a claim and prove it: $n^3+(n+1)^3>(n+2)^3$.

I think this statement holds when $n\ge6$, where $n^3+(n+1)^3=559$, and $(n+2)^3=512$.

I thought about using induction, but I couldn't express $(n+1)^3+(n+2)^3$ in terms of the inductive hypothesis- $n^3+(n+1)^3$.

So I'm wondering if there's a better way than induction to prove it or how to use the inductive hypothesis in a useful way given that induction is a good way to prove this.

Answer 1

For all natural $n<6$ it's wrong.

For all $n\geq6$ we have $$n^3+(n+1)^3-(n+2)^3=(n-6)(n^2+3n+9)+47>0.$$

Answer 2

Here is one way you can prove the result inductively:

Assume, for some integer $k\geq6$ it is true that $$k^3+(k+1)^3\gt(k+2)^3,$$ this implies $$(k+1)^3+(k+2)^3\gt2(k+2)^3-k^3.$$

Next, assume $$(k+3)^3\geq2(k+2)^3-k^3$$ and draw a contradiction (using the fact that $k\geq6$). This proves the case for $P(k+1)$.

Answer 3

Need to show:

$n^3\gt \dfrac{(n+2)^3-(n+1)3}{1}=$

$=3(t^2)$, $t \in (n+1,n+2)$ (MVT)

Hence:

$3(n+2)^2 \gt (n+2)^3-(n+1)^3$.

For $n \ge 6:$

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

Since:

The inequality is true for $n=6$.

$f(x):= x^3 -3(x+2)^2$.

$f'(x) = 3x^2 -6(x+2) =$

$3(x^2-2x -4) = $

$3[( x-1)^2 -5] >0$, for $x \ge 6$,

hence $f(x)$ is strictly increasing