I want to start by saying that I have far less trouble handling a non-inequality induction problem. I really don't understand the steps to take to get to the desired end product with these inequality induction proofs. That being said, I feel like I just wrote a mess on my paper that leads me nowhere. Here's my proof so far for the mathematical induction of $40(2n)!≥30^{n}$, where n ≥ 1
Let P(n) be the statement $40(2n)!≥30^{n}$ where n ≥ 1.
Basis Step: (n = 1) $LHS = 40 * 2! = 80$ and $RHS = 30^{1} = 30$
$80 ≥ 30$
Inductive Step: Assume $P(k)$ is true for $k = 1$. Our goal is to show $P(k+1)$ is true by showing $40(2(k+1))!≥30^{k+1}$ for $k ≥ 1$.
(Beyond this step I have no clue how to alter the LHS factorial or the RHS exponent in such a way to benefit me. Below is my work so far)
LHS: (I tried to multiply $(2k+2)(2k+1)$ to both sides, but then I didn't see how that would help)$40(2k)!≥30^{k} = (2k)!*40≥30^{k} = (2k+2)(2k+1)(2k)!*40≥30^{k}*(2k+2)(2k+1)$
RHS: $40(2(k+1))!≥30^{k+1}=(2k+2)!*40≥30^{k+1}=(2k+2)(2k+1)(2k)!*40≥30^{k+1}$ $=(2k+2)(2k+1)(2k)!*40≥30*30^{k}$
(I then assumed the inductive hypothesis and placed $40(2k)!≥30^{k}$ in the middle to get...)
$(2k+2)(2k+1)(2k)!*40≥40(2k)!≥30^{k}≥30*30^{k}$
At this point, I've got nothing. I know that $30^{k}≥30*30^{k}$ makes no sense, but I don't know where to move $30*30^{k}$ since I can't assume that $40(2k)!≥30*30^{k}$. I don't have any clue how I can manipulate either side to help me.
For $k \ge 2$, $(2k+1)(2k+2) \ge 30$. Hence if $P(k)$ is true for $k=1$ and $k=2$,
$$P(k+1) = 40(2k+2)! = 40(2k)! (2k+1)(2k+2)$$ $$\ge 30^k (2k+1)(2k+2)\quad\quad\quad (since\ 40(2k)! >= 30^k\ by\ P(k))$$ $$ \ge 30^k \cdot 30 \forall k \ge 2$$ $$ \ge 30^{k+1}$$
So you need to verify the proposition for $k=2$ and proceed with the induction.