# An inequality involving multi-index

I came across these inequalities while learning about Schwartz functions (Classical Fourier Analysis, Grafakos) and I have no idea how to prove this:

For $x \in \mathbb{R}^{n}$ and $\alpha = (\alpha_{1}, \ldots, \alpha_{n}) \in \mathbb{N}^{n}$, we set

$$x^{\alpha} = x_{1}^{\alpha_{1}}\cdots x_{n}^{\alpha_{n}}.$$

Then prove that there exists a constant $c_{n,\alpha}$ such that

$$\left| x^{\alpha}\right| \leq c_{n,\alpha}|x|^{|\alpha|}$$

where $|\alpha| = \alpha_{1} + \cdots + \alpha_{n}$.

Conversely, for every $k \in \mathbb{N}$, there exists a $C_{n,k}$ such that

$$|x|^{k} \leq C_{n,k}\sum\limits_{|\beta| = k}|x^{\beta}|$$

Any help would be appreciated.

P.S. Please let me know if the question is too elementary for this forum.

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What is $|x|$? The Euclidean or $l_2$ norm? – Todd Trimble Jul 28 '13 at 16:04
$|x| = \sqrt{x_{1}^{2} + \cdots + x_{n}^{2}}$ – Vishal Gupta Jul 29 '13 at 2:50

The first inequality with constant 1 follows from $\vert x^\alpha\vert\le\Vert x\Vert_{\infty}^{\vert \alpha \vert},\quad\text{where$\Vert x\Vert_{\infty}$is the sup-norm.}$
The second equality, also with constant 1, is due to $\Vert x\Vert_{\infty}^k=\max_{1\le j\le n} \vert x_j\vert^k\le \max_{\vert \alpha\vert=k} \vert x^\alpha\vert.$