Prove $\varphi\in\mathcal{S}(\mathbb R^n)$ if and only if the following inequality holds.. I need some help for showing the following result: Let $\varphi\in C^\infty(\mathbb R^n)$. Then $\varphi\in \mathcal{S}(\mathbb R^n)$ if and only if for all $\alpha\in\mathbb N^n$ and $N\geq 0$ there is a constant $C_{\alpha, N}$ such that $$|\partial^\alpha \varphi(x)|\leq \frac{C_{\alpha, N}}{(1+|x|)^N}.$$ 
For me:
$C^\infty_c(\mathbb R^n)$ is the set of all $\varphi:\mathbb R^n\rightarrow \mathbb C$ of class $C^\infty$.
$\mathcal{S}(\mathbb R^n)$ is the Schwartz space defined as the class of all $C^\infty$ functions $\varphi:\mathbb R^n\rightarrow \mathbb C$ such that $$\sup_{x\in\mathbb R^n}|x^\beta \partial^\alpha \varphi(x)|<\infty,$$ for all multi-indices $\alpha, \beta\in\mathbb N^n$.
I proved the implication $(\Rightarrow)$ as follows: Given $\alpha, \beta\in\mathbb N^n$, $$|x^\beta \partial^\alpha \varphi(x)|=|x^\beta||\partial^\alpha \varphi(x)|\leq |x|^{|\beta|}|\partial^\alpha \varphi(x)|\leq (1+|x|)^{|\beta|}\frac{C_{\alpha, |\beta|}}{(1+|x|)^{|\beta|}}=C_{\alpha, |\beta|},$$ hence $\displaystyle\sup_{x\in\mathbb R^n}|x^\beta \partial^\alpha \varphi(x)|<\infty$, i.e., $\varphi\in\mathcal{S}(\mathbb R^n)$.
The main problem is the converse. I conjecture the proof must be made by contradiction but I wasn't managed to do it.. 
Any help with be great..Thanks 
 A: If $\varphi\in\mathcal S(\mathbb R^n)$, then $x^{\alpha}\partial^{\beta}\varphi$ and $P(x)Q(\partial)\varphi$  are  in $\mathcal S(\mathbb R^n)$, for all polynomials $P$, $Q$ and multi-indices $\alpha$, $\beta$. Then, selecting any polynomial $P=P(x)
$ of degree $k$ and $Q(\partial)=\partial^{\beta}$ we can write
$$|P(x)\partial^{\beta}\varphi(x)|\leq constant_{k,\beta}; $$
taking $P(x)=(1+x^2_1+\dots+x^2_n)^k$ we arrive at the thesis.
A: Using @Avitus idea I wrote the following proof for the converse:
On the other hand, given $\alpha=(\alpha_1, \ldots, \alpha_n)\in\mathbb N^n$, $$|(1+\sum_{i=1}^nx_i^2)^N\partial^\alpha \varphi(x)|=|(1+\sum_{i=1}^nx_i^2)^N||\partial^\alpha\varphi(x)|.$$ But by the multinomial theorem,
$$(1+\sum_{i=1}^nx_i^2)^N=\sum_{|(\beta_0, \beta)|=N}\frac{N!}{\beta_0!\beta!}x^{2\beta}, \beta_0\in\mathbb N.$$ Notice both $\beta_0$ and $\beta$ depend on $N$. Hence, $$|(1+\sum_{i=1}^nx_i^2)^N\partial^\alpha \varphi(x)|\leq \sum_{|(\beta_0, \beta)|=N}\frac{N!}{\beta_0!\beta!}|x^{2\beta}\partial^\alpha \varphi(x)|\leq \sum_{|(\beta_0, \beta)|=N}\frac{N!}{\beta_0!\beta!}\sup_{x\in\mathbb R^n}|x^{2\beta}\partial^\alpha \varphi(x)|=C_{\alpha, N}.$$ The result follows noting that $$(1+|x|)^N\leq (1+\sum_{i=1}^nx_i^2)^N.$$ 
The above argument works well if $N\in\mathbb Z^+$ otherwise we can't use the multinomial theorem.. 
