Inequality Proof – Prove n^2 > (n+1) for All Integers n ? 2

Question

I understand that I need to use induction for this, that's not a problem. I get stuck after I try to invoke the inductive hypothesis.

$P_n: n^2 > n+1$… and we want to prove $P_{n+1}: (n+1)^2 > (n+1)+1$ holds.

Base: $P_2: 4 > 3$

Suppose $P_n$ holds, such that $n^2 > n+1$.

Now, in most of the induction proofs I've done have some manipulation of the original formula to get the new step into form. I don't necessarily see where to go with this.

My original thought was, if $P_{n+1}$ is different from $P_n$ by one unit, then I'd add a unit to the other side of the formula, and get it into form. This turns out to be difficult.

Second thought was to start by looking at $(n+1)^2 = n^2+2n+1$ and since there's an $n^2$ term, I could use the "inductive hypothesis", where I isolate the squared term, and check to see if the RHS of the inequality is less than the RHS in the inductive step………. This doesn't seem right either.

Thanks for the advice

Answer 1

Do you really need induction?

$$0 < (n-1)^2 = n^2 - 2n + 1 \implies n^2 > 2n - 1 = n + (n-1) \ge n + 1$$