Prove that $(\alpha_1+\cdots+\alpha_m)!\le\prod_{j=1}^m\alpha_j! (2j)^{2\alpha_j}$

Question

When I am reading the book "Stochastic Partial Differential Equations. A Modeling, White Noise Functional Approach(Second Edition)", I have a problem proving the inequality that $$n! \le \alpha!(2\mathbb{N})^{2\alpha}$$ In this formula, $\alpha=(\alpha_1,\cdots,\alpha_m)$ is a vector satisfying that $\alpha_1+\cdots+\alpha_m=n$. In the previous pages, we have the definitions that $\alpha!=\alpha_1!\alpha_2!\cdots\alpha_m!$ and $(2\mathbb{N})^\alpha=\prod_{j=1}^m (2j)^{\alpha_j}$. So the problem is equivalent to prove that $$(\alpha_1+\cdots+\alpha_m)!\le\prod_{j=1}^m\alpha_j! (2j)^{2\alpha_j}$$ Actually, there's a method given by the book, but I think there's a mistake about the proof. Can anyone solves this problem? Thanks a lot!

Answer 1

We can prove by induction though.

The base case is trivial. Now we assume for $n-1$ it is correct.

We try to decrease one of $\alpha_1,\dots,\alpha_m$ by $1$. If we decrease $\alpha_i$ by $1$, the left hand side is shrink by $1/n$, and the right hand side is going to shrink $1/(\alpha_i (2i)^2)$. Assume all of these $1/(\alpha_i (2i)^2)$ is greater than $1/n$. In other words, $\alpha_i (2i)^2\le n$ for all $i$. So, we have $\alpha_i\le n/(2i)^2$. Summing those, we have $\alpha_1+\dots+\alpha_m\le n/(2^2)+n/(4^2)+n/(6^2)+\dots=n\times \pi^2/24<n,$ which is a contradiction. So, we can find some $i$ such that $1/(\alpha_i (2i)^2)<1/n$ or $(2i)^2\alpha_i>n$. The induction hypothesis claims that

$$(\alpha_1+\dots+\alpha_{i-1}+(\alpha_i-1)+\alpha_{i+1}+\dots+\alpha_n)!\le\prod_{j=1,j\ne i}^m\alpha_j!(2j)^{2\alpha_j}\times (\alpha_i-1)!(2i)^{2\alpha_i-2}$$

Multiply $n$ on the left hand side and $(2i)^2\alpha_i$ on the right hand side yields the result.

Answer 2

I've already solved this problem using the induction. Later, I will post my answer here.

To prove $n!\le\prod_{j=1}^m(2j)^{2\alpha_j}\alpha_j!$, where $n=\alpha_1+\cdots+\alpha_m, \forall m$

Using the inducing method, when $n=1$, it is obvious that $$1\le n!\le\prod_{j=1}^m(2j)^{2\alpha_j}\alpha_j!=(2m)^{2\alpha_m}=(2m)^2$$

Now, if it is true for $n$, then we need to prove it is also true for $\alpha_1+\cdots+\alpha_m=n+1$.

Notice that there exists $i$ such that $4i^2\alpha_i\ge n+1$. Otherwise, if for any $i$, there is $4i^2\alpha_i< n+1$, then we have $$\frac{1}{4i^2}>\frac{\alpha_i}{n+1}, \forall i.$$ Sum this up through $i$ yield that $$\sum_{i=1}^m \frac{1}{i^2}>4\sum_{i=1}^m\frac{\alpha_i}{n+1}=4,$$ which is not true.

Then let $\beta_j=\alpha_j$ for $j\ne i$ and $\beta_i=\alpha_i-1$. Then, we have $$(n+1)!=n!(n+1)\le(n+1)\prod_{j=1}^m\beta_j!(2j)^{2\beta_j}=\prod_{j\ne i}\alpha_j!(2j)^{2\alpha_j}\cdot(\alpha_i-1)!(2i)^{2\alpha_i}(n+1)$$

Because of the property of $i$, we have $$(\alpha_i-1)!(2i)^{2(\alpha_i-1)}(n+1)\le(\alpha_i-1)!(2i)^{2\alpha_i}\frac{1}{4i^2}4i^2\alpha_i=\alpha_i!(2i)^{2\alpha_i}$$ Thus the desired result has been proved

Answer 3

We use two facts $\frac{\pi^2}{6}=\sum_{j=1}^{\infty}\frac{1}{j^2}\equiv 1,65$ and $$\frac{1}{2\pi}\int_0^{2\pi}e^{inx}dx=1\ \mathrm{for}\ n=0,\ \mathrm { and }\ 0\ \mathrm{for}\ n=0.\ \ \ (*)$$ With obvious notations, if $(\sum_{j=1}^m\frac{t_j}{4j^2})^n=\sum_{\alpha}a_{\alpha}t^\alpha,$ we asked to prove that $a_{\alpha}<1.$ Denote $\langle\alpha,x\rangle=\alpha_1x_1+\cdots+\alpha_mx_m.$ Consider the function $$f(x)=f(x_1,\ldots,x_m)=\sum_{j=1}^m\frac{e^{ix_j}}{4j^2}$$ which satisfies $|f(x)|\leq \frac{\pi^2}{24}.$ Finally from (*) we have$$a_{\alpha}=\frac{1}{(2\pi)^m}\int_0^{2\pi}\ldots\int_0^{2\pi}(f(x))^ne^{-i\langle\alpha,x\rangle}dx_1\ldots dx_m$$$$\leq \frac{1}{(2\pi)^m}\int_0^{2\pi}\ldots\int_0^{2\pi}|f(x)|^ndx_1\ldots dx_m\leq \left(\frac{\pi^2}{24}\right)^n<1.$$