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Would you please check if my answer is correct and confirm that it is proof by induction? Thank you.

The proof is by induction.

Base Case: when $n=1$:

$$\begin{eqnarray}1^{3}&>&1^{2}-6(1)+4\\ 1&>&-1\end{eqnarray} $$

Thus, the base case is proven.

Inductive Step: suppose $n^{3}>n^{2}-6n+4$ and $n\ge2 , n\in\mathbb{N}$ .

$$ \begin{eqnarray} n^{3}+(3n^{2}+3n+1) & > & n^{2}-6n+4+(3n^{2}+3n+1) \\ n^{3}+3n^{2}+3n+1 & > & 4n^{2}-3n+5 \\ n^{3}+3n^{2}+3n+1 & > & n^{2}+2n+1-6n-6+4+3n^{2}+n+6 \\ (n+1)^{3} & > & (n+1)^{2}-6(n+1)+4+3n^{2}+n+6 \\ & > & (n+1)^{2}-6(n+1)+4 \\ \end{eqnarray} $$ since $n\ge2 , n\in\mathbb{N} $.

This proves the inductive step, so the result follows by induction.

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Overall correct. Style is high school two column, most lines, starting with $1^3>1^2-6(1)+4$. Should change, unless you are in high school. Small almost typo, want $n\ge 1$. Else in principle have excluded step from $1$ to $2$. I prefer going from $n=k$ to $n=k+1$. – André Nicolas May 2 '12 at 17:48

Here is a simpler, more conceptual way, by telescopy. First do the obvious inductive proof of the lemma that $\rm\:f(n) > 0\:$ for $\rm\:n\ge 1\:$ if $\rm\:f(1)> 0\:$ and $\rm\:f\:$ is increasing: $\rm\: f(n+1) \ge f(n)\:$ for $\rm\:n\ge1.\:$

So for $\rm\:f(n) = n^3\!-n^2+6n\!-4\:$ it suffices to show $\rm\: f(n\!+\!1)-f(n) = 4n^2+2n+6 \ge 0\:$ for $\rm\:n\ge 1,\:$ which is easy. I have not checked the details of your proof, but it can be rearranged to have the above telescopic form, an exercise that is instructive (and helpful for proof checking).

Note: the proof may be viewed as reducing the induction to the trivial induction that a sum of positive terms is positive, since by telescopy we may represent $\rm\:f\:$ as the sum of its first differences $\rm\:f(n) = f(1) + \sum_{k=1}^{n-1}\: (f(k+1)-f(k)),\:$ where $\rm\:f(k+1)-f(k) \ge 0.\:$ Follow the link for more.

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