Show that n^3 (n+1)^2 for all natural numbers n=3

Question

Show that n^3 > (n+1)^2 for all natural numbers n>=3

ganeshie8 · Answer

$n\ge 3$

multiply both sides by $n^2$ and get
$n^3\ge 3n^2\=n^2+n^2+n^2 \=n^2 + n\cdot n+n^2 \\gt n^2 + 2\cdot n + 1 \= (n+1)^2$

mathmate · Answer

@ganeshie8  : brilliant!

Or if we prefer the long way, using mathematical induction,

Statement: $n^3>(n+1)^2$  for n>=3.

Base case: for n=3, we have
$n^3-(n+1)^2=3^3-4^2=11 > 0$ so statement holds for n=3.

Inductive Hypothesis: assuming statement holds for case n=k.
i.e. $k^3-(k+1)^2 = S_k  >0$

Inductive step: for case n=k+1, we have
$(k+1)^3-(k+1+1)^2 =k^3+3k^2+3k+1-(k^2+4k+4)$
$=[k^3-(k^2+2k+1)] + 3k^2+k-2)$
$=S_k +(3k-2)(k+1)$
On the last line, 
$S_k >0 $  by hypothesis
(3k-2)   > 0  $\forall k\ge3$
(k+1)     >0   $\forall k\ge3$
Therefore  $=S_k +(3k-2)(k+1) > 0$  for case n=k+1

Conclusion:
By mathematical induction, we have established that statement $n^3>(n+1)^2$  is true for all integers n>=3

jhannybean · Answer

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