How do I solve $c^2 + 1 > 2c - 1$ when $c \ge 2$?

Question

We have been given a question to do as part of a task set by our lecturer, however he has failed to provide any examples of similar questions with all the corona stuff going on, and I cannot find the name of the method used to solve it. The question is as follows.

Suppose that $c\in \Bbb N$. Prove that: If $c\ge 2$ then $c^2+1>2c−1$.

What is this type of question called, and what would be the working to solve it?

Answer 1

This is a good example of when proof by induction is useful. The base case is $c=2$. Then assume that the inequality holds for some $k>2$ and show that the inequality holds for $k+1$.

Answer 2

$$(c-1)^2\geq 0>1\Rightarrow c^2-2c+1> -1\Rightarrow c^2+1>2c-1$$ Now this would be the case for any $c\in N$, thus valid for $c\geq 2$.

(I think you meant some else equation)