calculate lower bound of busy beaver
All you need to do is show that a machine with $m$ states can halt with $\Omega(2^m)$ 1's on the tape. 

There are many ways to prove this. This is just one suggestion, and it may be sub-optimal.

Let $T$ be a machine that on a word of the form $0^i1^j$, when the head is in the leftmost $0$, halts with $0^i1^{2j}$ on the tape and the head at the leftmost $0$.

$T$ is a fixed machine, and thus has $t$ states, for a constant $t$.

We now construct the machine $M_i$ as follows: first, $M_i$ writes $0^i1$ on the tape, using $i+1$ states. It then moves the head to the left most cell, using another single state.
Then, if the leftmost cell is $0$, it erases it and passes control to $T$ (this requires $t+1$ states).
If the leftmost cell is $1$, it halts.

$M_i$ has $i+2+t$ states, and after it halts the tape contains exactly $2^i$ 1's (since in every call to $T$ we double the number of 1's, and we call $T$ $i$-times).

Thus, if $M_i$ has $m$ states, it halts with $\frac{2^m}{t+2}\in\Omega(2^m)$ 1's, and we are done.