Prove by induction: $2!\cdot 4!\cdot 6!\cdot\cdot\cdot (2n)!\ge ((n+1)!)^n$

Question

Prove by induction: $2!\cdot 4!\cdot 6!\cdot\cdot\cdot (2n)!\ge ((n+1)!)^n$

For $n=1$ inequality holds.

$(*)$For $n=k$

$2!\cdot\cdot\cdot (2k)!\ge ((k+1)!)^k$

Multiplying LHS and RHS with $(2k+2)!$ gives

$$2!\cdot\cdot\cdot (2k)!(2k+2)!\ge ((k+1)!)^k(2k+2)!$$

Assume (by contradiction)$$2!\cdot\cdot\cdot (2k)!(2k+2)!< ((k+1)!)^k(2k+2)!$$ $$2!\cdot\cdot\cdot (2k)!(2k+2)!-((k+1)!)^k(2k+2)!<0$$ $$(2k+2)!(2!\cdot\cdot\cdot (2k)!-((k+1)!)^k)<0$$

$(2!\cdot\cdot\cdot (2k)!-((k+1)!)^k)\ge 0$ by $(*)$, thus inequality holds $\forall n\in\mathbb{N}$

Is this proof correct?

Answer 1

First, show that this is true for $n=1$:

$\prod\limits_{k=1}^{1}(2k)!\geq(1+1)!^{1}$

Second, assume that this is true for $n$:

$\prod\limits_{k=1}^{n}(2k)!\geq(n+1)!^{n}$

Third, prove that this is true for $n+1$:

$\prod\limits_{k=1}^{n+1}(2k)!=$

$\color\red{\prod\limits_{k=1}^{n}(2k)!}\cdot(2(n+1))!\geq$

$\color\red{(n+1)!^{n}}\cdot(2(n+1))!=$

$(n+1)!^{n}\cdot(2n+2)!=$

$(n+1)!^{n}\cdot(n+1)!\cdot\dfrac{(2n+2)!}{(n+1)!}=$

$(n+1)!^{n+1}\cdot\dfrac{(2n+2)!}{(n+1)!}=$

$(n+1)!^{n+1}\cdot\underbrace{(n+2)\cdot(n+3)\cdot\ldots\cdot(2n+2)}_{n+1\text{ times}}\geq$

$(n+1)!^{n+1}\cdot\underbrace{(n+2)\cdot(n+2)\cdot\ldots\cdot(n+2)}_{n+1\text{ times}}=$

$(n+1)!^{n+1}\cdot(n+2)^{n+1}=$

$(n+2)!^{n+1}$

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Please note that the assumption is used only in the part marked red.

Answer 2

Assume (by contradiction) $$2!⋅⋅⋅(2k)!(2k+2)!<((k+1)!)k(2k+2)!$$

Why would you assume that, you just showed the contrary one line above ?

Once you're here $$2!\cdot\cdot\cdot (2k)!(2k+2)!\ge ((k+1)!)^k(2k+2)!$$ you just have to show that $((k+1)!)^k(2k+2)! \ge (k+2)!^{k+1}$.

$(k+2)!^{k+1} = (k+1)!^k \times (k+2)^k \times (k+2)!$ and $(k+2)^k \times (k+2)! \le (2k+2)!$ (because $k+2 < k+2+i$ for all $i$). Thus, you have your result.

Answer 3

Define $S$ the wanted number. Then $$ S=\prod_{i=1}^n (2i)!=\prod_{i=1}^n (2(n+1-i))!. $$ Therefore, based on $a!b!\ge \max(a,b)!\ge \frac{a+b}{2}!$ whenever $a,b$ are integers with same parity, we conclude that $$ S=\sqrt{\prod_{i=1}^n(2(n+1-i))!(2i)!}\ge \sqrt{\prod_{i=1}^n(n+1)!^2}=(n+1)!^n. $$

Answer 4

That is equivalent to: $$ \sum_{k=1}^{n} \log\Gamma(2k+1) \geq n \log\Gamma(n+2) \tag{1}$$ that is a trivial inequality, since $\log\Gamma$ is a convex function, due to the Bohr-Mollerup theorem, for instance.