Proof verification for $(2n)! ≥ 2^n (n!)^2$

Question

Can someone check my proof for $(2n)! ≥ 2^n (n!)^2$ for positive integers?

Base case $$\begin{align*}(2\cdot1)! &\geq 2^1 (1!)^2,\\ 2 &\geq 2\end{align*}$$ true for $n=1$
Assume true for $n=k:$ $$(2k)! \geq 2^k (k!)^2.$$ Let $n = k+1.$ $$(2k+2)! ≥ 2^{k+1} \times ((k+1)!)^2.$$ By assumption, $$\begin{align*}(2k)! &\geq 2^k \times (k!)^2\\ (2k+2)! &\geq ((2^k)(k!)^2)(2k+1)(2k+2)\\ &\geq (2^k \times (k!)^2)(4k^2 + 6k +2)\\ &\geq (2^k \times (k!)^2)(12)&\text{since $k ≥ 1$}\\ &\geq 2^{k+1} \times 6 \times (k!)^2\\ &\geq 2^{k+1} \times 6 \times((k+1)(k!)/(k+1))^2\\ &\geq 2^{k+1} \times 6 \times ( (k+1)!/2)^2 &\text{since $k ≥1$}\\ &\geq 2^{k+1} \times 1.5 ((k+1)!)^2\\ &\geq 2^{k+1} \times ((k+1)!)^2.\end{align*}$$ Therefore true for $n=k+1.$

I've improved the mathjax. I think it is all correct, but please check to see that no extra errors have crept in. Also, that $10$ should be a $6$, as $12=2\times6$ not $2\times10$. — user1729, May 24 '23 at 08:23
Does this answer your question? Prove that $2^n(n!)^2 \leq (2n)!$ - found using an Approach0 search. There are also other duplicates here, e.g., Use mathematical induction to prove $ (2n)!\geq 2^n(n!)^2$ for $n \in \mathbb{N}$, Question on mathematical induction which involves factorial and inequality., etc., as well as duplicate AoPS threads, e.g., ... — John Omielan, May 24 '23 at 08:39
(cont.) a very easy inequality, inequality, etc. Finally, welcome to Math SE. — John Omielan, May 24 '23 at 08:42
However, for a solution-verification question, you must specify precisely which step in the proof you question, and why so. This site is not meant to be used as a proof checking machine. — Anne Bauval, May 24 '23 at 08:50
@JohnOmielan Since it was explicitely a solution-verification question, it cannot be closed as a duplicate, can it? — Anne Bauval, May 24 '23 at 09:01
@AnneBauval This question obviously can be closed as a duplicate, so I assume you're asking if it's appropriate. I've read various meta posts which, to varying degrees, deal with this issue. As far as I can tell, there's no specific consensus. IMHO, unless the proposed solution is particularly novel, so this is usually especially applicable for relatively simple proofs like this one, the proposed duplicate(s) use a similar enough approach to be considered a duplicate. I believe that is the case here, but I'll leave it up to you, & the other members, to decide for themselves. — John Omielan, May 24 '23 at 09:14
@AnneBauval Nonetheless, if there's a specific issue or problem with what the OP is asking about (especially if they do as you requested, i.e., point out where they're unsure of what they're doing), then if it's simple enough, then it can be pointed out in a comment, or perhaps in an answer before the question is then potentially closed as a duplicate. — John Omielan, May 24 '23 at 09:17
This proof is simple indeed but mistaken, so that the (second part of the) answer below is more appropriate than considering it a duplicate. But I agree that a comment would have been sufficient to point this mistake. — Anne Bauval, May 24 '23 at 09:18

score 2 · Answer 1 · answered May 24 '23 at 08:37

I would do it like this: $$(2n)!\ge2^n(n!)^2$$ $$\frac{(2n)!}{n!}\ge2^n\cdot n!$$

Notice that...

$\frac{(2n)!}{n!}=\color{#AA0000}{2n}\color{#AA7700}{(2n-1)}\color{#AAAA00}{(2n-2)}\dots\color{#77AA00}{(n+2)}\color{#00AA00}{(n+1)}$, which has $n$ elements
$n!=\color{#AA0000}{n}\color{#AA7700}{(n-1)}\color{#AAAA00}{(n-2)}\dots\color{#77AA00}2\cdot\color{#00AA00}1$, which also has $n$ elements

We can pair those two together like this: $$\color{#AA0000}{2n}\ge2\cdot \color{#AA0000}n$$ $$\color{#AA7700}{2n-1}\ge2\color{#AA7700}{(n-1)}$$ $$\color{#AAAA00}{2n-2}\ge2\color{#AAAA00}{(n-2)}$$ $$\dots$$ $$\color{#77AA00}{n+2}\ge2\cdot\color{#77AA00}2$$ $$\color{#00AA00}{n+1}\ge2\cdot\color{#00AA00}1$$

So, when we multiply them together: $$2n(2n-1)\dots(n+2)(n+1)\ge2^n\cdot n!$$ $$\frac{(2n)!}{n!}\ge2^n\cdot n!$$ $$(2n)!\ge2^n(n!)^2$$ Q.E.D. (cool Latin phrase meaning "proven")

Also, about your proof, I think it's slightly wrong to go from $$(2k+2)!\ge(2^k\cdot(k!)^2)(4k^2 + 6k +2)$$ $$\text{to}$$ $$(2k+2)!\ge(2^k\cdot(k!)^2)(12)\;\text{since}\;k ≥ 1$$

Hmm, why's that? It's because $k$ can be more than $1$, in which case $4k^2+6k+2$ would have a greater value.

Let me show you an example false proof

Hi! Today, I will prove that $x^2\ge\frac12x^3$ for $x\ge1$. $$x^2\ge\frac12x^3$$ $$x^2\ge\frac12\;\text{since}\;x\ge1$$ $$x\ge\sqrt\frac12$$ $$x\ge\frac12\sqrt2$$ Because $x$ can't be less than $1$, it is proven that $x^2\ge\frac12x^3$. Yeay!

A shorter (and slightly simpler) version of your proof (which imo is anyway off topic since the question was proof-verification) was already available in the first link given above by John. (+1) however, for the second part of the answer. — Anne Bauval, May 24 '23 at 09:08

Proof verification for $(2n)! ≥ 2^n (n!)^2$

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