Stuck trying to prove an inequality regarding the product of factorials

Question

I am to prove by induction that $$2!\cdot4!\cdot6!\cdots(2n)!\ge[(n+1)!]^n$$ I've shown the base case, assumed it is true and then tried proving it for $n+1$. However I am unable to do this. I have tried two things. First after writing the expression for $n+1$ $$2!\cdot4!\cdot6!\cdots(2n)!\cdot(2n+2)!\ge[(n+2)!]^{n+1}$$ I've tried using the assumption directly to obtain $$(2n+2)!\ge(n+2)^{n+1}(n+1)!$$ however that is not true for obvious reasons. I've also tried multiplying the assumption inequality with $(n+2)^n(n+2)!$ to obtain $$2!\cdot4!\cdot6!\cdots(2n)!(n+2)^n(n+2)!\ge[(n+2)!]^{n+1}$$ and then tried to prove that $$2!\cdot4!\cdot6!\cdots(2n)!\cdot(2n+2)!\ge2!\cdot4!\cdot6!\cdots(2n)!\cdot(n+2)!(n+2)^n$$ but again I end up at $$(2n+2)!\ge(n+2)^{n+1}(n+1)!$$ which can't be true. Any help?

Answer 1

Your proof is perfectly OK up to the point when you say

$$(2n+2)!\ge(n+2)^{n+1}(n+1)!$$ which can't be true.

In fact, this inequality can be true and it is.

---

You can show this inequality by seeing that $$(2n+2)! = (2n+2)\cdot (2n+1)\cdots (n+2) \cdot (n+1)!$$

Answer 2

First note that $(2n+2)!\ge(n+1)!(n+2)^{n+1}$ because all $n+1$ factors from $n+2$ to $2n+2$ exceed $n+2$.

Then by the induction hypothesis,

$$2!\cdots(2n)!(2n+2)!\ge(n+1)!^{n}(2n+2)!\ge(n+1)!^n(n+1)!(n+2)^{n+1}=(n+1)!^{n+1}(n+2)^{n+1}=(n+2)!^{n+1}.$$