My question concerns on the one hand a specific inequality and on the other hand a general strategy on how to approach inequalities in general.

Usually I don't have problems using induction in order to prove equalities. But wih inequalities my brain just stops working.

Given $n^2-2n-1>0$ for $n>3$.

I would like to prove that via induction from $n-1$ to $n$.

So $(n-1)^2-2(n-1)-1=n^2-2n+1-2n+2-1=n^2-4n+2$.

From here on I just don't have a clue...


1 Answer 1


First you prove the base case: clearly for $n=4$ you have $4^2-2\cdot 4-1>0$.

So you assume it holds for $n-1$ (where $n>4$) and then prove it holds for $n$.

So you're assuming that $n^2-4n+2>0$.

You want to prove that then $n^2-2n-1>0$.

Well, this is clear, since $n^2-2n-1>n^2-4n+2,\forall n>4$.

However, induction here proves $n^2-2n-1>0,\forall n>3$ with $n$ integer. This doesn't prove that it holds for all real numbers in $(3;+\infty)$.

Alternatively, $n^2-2n-1=(n-1)^2-2$. Here we even more clearly have that if $(n-1)^2-2>0,n>3$, then $n^2-2>(n-1)^2-2>0$.


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