Running time of algorithm A is at least O(n²) - Why is it meaningless?

Question

Why is the statement:

The running time of algorithm A is at least O(n²)

is meaningless ?

The running time of Insertion sort algorithm is at most O(n²)

Is it Correct?

I tried the net but could not get a good explanation.

I have another question:

I know that any linear function a⋅n+b is O(n) and also O(n²). Is it also O(n³)?

Answer 1

T(n): running time of Algo A. We just care about the upper bound and lower bound of T(n) The statement: T(n) >= O(n^2)

Upper bound: Because T(n) >= O(n^2), then there's no information about upper bound of T(n) Lower bound: Assume f(n) = O(n^2), then the statement: T(n) >= f(n), but f(n) could be anything that is "smaller" than n^2 , Ex: constant, n,..., So there's no conclusion about lower bound of T(n) too

=> The statement is meaningless