Prove that $\forall n\ge2\in \mathbb{N}$ ${\prod\limits_{j=1}^{n}(2j)!}>[(n+1)!]^n$

Question

How would you prove this? Would you use induction?

Prove that $$\forall n\ge2\in \mathbb{N}$$

$${\prod\limits_{j=1}^{n}(2j)!}>[(n+1)!]^n$$

Answer 1

You can prove it by direct computation. Show

$${\prod\limits_{j=1}^{n}(2j)!}>[(n+1)!]^n$$

Write what you have on the LHS and RHS:

$$ (1 \cdot 2) \cdot (1 \cdot 2 \cdot 3 \cdot 4) \cdots > 1^n \cdot 2^n \cdots (n+1)^n$$

$$ 1^n \cdot 2^n \cdot 3^{n-1} \cdot 4^{n-1} \cdots (2 j -1)^{n+1-j} \cdot (2 j)^{n+1-j} \cdots (2 n - 1) \cdot (2 n) > 1^n \cdot 2^n \cdots (n+1)^n$$

Now divide by all terms up to base (n+1):

Let $n$ be even (odd $n$ works equivalently): $$ (n+2)^{n/2} \cdot (n+3)^{n/2 -1} \cdot (n+4)^{n/2 -1} \cdots (2 n - 1) \cdot (2 n) > 3^{1} \cdot 4^{1} \cdot 5^2 \cdot 6^2 \cdots (n+1)^{n/2} $$

There are $(n-1)$ terms on both sides. Now we compare pairs of terms with the same exponent on the LHS and RHS and it is evident that in each of these pairs, the base on the LHS is always larger than the base on the RHS. This proves the claim.