Proving inequality using Mathematical Induction

Question

I'm stuck in a prove by Mathematical Induction and I have no clue how to continue on.
To prove:

$2!\cdot 4! \cdot 6! \cdot ...\cdot(2n)! \geq [(n+1)!]^n$ for all $n \in \mathbb{N}$

Proof:
We proceed by induction. Since $2! = 2 \geq 2^1 =2$ is true for $n = 1$. Assume for a $k \in \mathbb{N}$, that
$2!\cdot 4! \cdot 6! \cdot ...\cdot(2k)! \geq [(k+1)!]^n$ .
We show that
$2!\cdot 4! \cdot 6! \cdot ...\cdot(2(k+1))! \geq [(k+2)!]^{k+1}$

Observe that
$2!\cdot 4! \cdot 6! \cdot ...\cdot (2k)! \cdot (2(k+1))! \geq [(k+1)!]^k (2k+2)!$

I do not, however, know how to proceed further.

Answer 1

We wish to show $2!4!\cdots (2k+2)! \geq [(k+2)!]^{k+1}$. By the inductive hypothesis

$$[(k+2)!]^{k+1}=[(k+1)!]^k\cdot (k+1)!(k+2)^{k+1}\leq 2!4!\dots(2k)!\cdot(k+1)!(k+2)^{k+1}.$$

So, the result holds if we can show that

$$(k+1)!(k+2)^{k+1}\leq (2k+2)!$$

However, we can estimate $(k+2)$ by $(k+n)$ for all $n\geq 2$:

$$(k+1)!(k+2)^{k+1}\leq(k+1)!(k+2)(k+3)\cdots (k+(k+2)) = (2k+2)!$$

and the result follows.

Answer 2

Expand out $[(k+2)!]^{k+1}$ into individual factors: $[1 \cdot 2 \cdot \ldots \cdot k \cdot (k + 1) \cdot (k + 2)]^{k+1} = 1^{k+1} \cdot 2^{k+1} \cdot \ldots \cdot k^{k+1} \cdot (k+1)^{k+1} \cdot (k+2)^{k+1}$. Doing the same thing with $[(k+1)!]^k$, we get $1^k \cdot 2^k \cdot \cdots \cdot k^k \cdot (k+1)^k$. Notice that for the numbers $1$ through $k+1$, we have exactly one more factor in $[(k+2)!]^{k+1}$ than $[(k+1)!]^k$. Grouping together everything we can, we can write $[(k+2)!]^{k+1} = [(k+1)!]^k \cdot 1 \cdot 2 \ldots \cdot k \cdot (k+1) \cdot (k+2)^{k+1}$. Simplifying a little, $[(k+2)!]^{k+1} = [(k+1)!]^k \cdot (k+1)! \cdot (k+2)^{k+1}$.

So now the question is: how do you know that $(2k+2)! > (k+1)! \cdot (k+2)^{k+1}$? For that, try the same trick: expand out both sides, cancel out factors, and see what happens.

Answer 3

Assume that $2!\cdot 4! \cdot 6! \cdot ...\cdot(2k)! \geq [(k+1)!]^n$, then $$ \begin{align} [(k+2)!]^{k+1} = & [(k+2)!]^{k}\cdot (k+2)!\\ = & [(k+1)!]^{k}\cdot(k+2)^{k}\cdot (k+2)! \\ \leq & [(k+1)!]^{k} \cdot \prod_{i=1}^k (k+2+i)\cdot (k+2)!\\ =& [(k+1)!]^{k} \cdot (2k+2)! \\ \leq & 2!\cdot 4! \cdot 6! \cdot ...\cdot (2k)! \cdot (2(k+1))! \end{align} $$