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.
 A: The first inequality is clear at $x=0$ with any constant . Now note that it suffices to prove it for $|x|=1$. This is because if it is true on the unit sphere then since $x\neq0$ then $y=(\frac{x_{1}}{|x|},\frac{x_{2}}{|x|},...,\frac{x_{n}}{|x|})=\frac{x}{|x|}\in S^{n-1}$ and then we have 
$$\begin{align} \frac{|x^{\alpha}|}{|x|^{|\alpha|}} 
& = \frac{ \sqrt{ x_1^{2\alpha_1} + \dots + x_n^{2\alpha_n} } }{|x|^{|\alpha|} }  \\
& = \bigg( \frac{ x_1^{2\alpha_1} + \dots + x_n^{2\alpha_n} }{|x|^{2|\alpha|}} \bigg)^{\frac{1}{2}}
\\
& = \bigg( \frac{x_1^{2\alpha_1}}{|x|^{2|\alpha|}} + \dots + \frac{x_n^{2\alpha_n}}{|x|^{2|\alpha|}} \bigg)^{\frac{1}{2}}
\\ 
& \leq \bigg[ \bigg(\frac{x_1}{|x|}\bigg)^{2\alpha_1} + \dots + \bigg(\frac{x_n}{|x|}\bigg)^{2\alpha_n}\bigg]^{\frac{1}{2}}
\\
& = \bigg| \bigg( \frac{x_1^{\alpha_1}}{|x|^{\alpha_1}}, \dots, \frac{x_n^{\alpha_n}}{|x|^{\alpha_n}} \bigg) \bigg| 
\\
& = | y^{\alpha} | 
\\
& \leq c_{n,\alpha} |y|^{|\alpha|}
\\
& = c_{n,\alpha}\bigg|\bigg(\frac{x_{1}}{|x|},\frac{x_{2}}{|x|},...\frac{x_{n}}{|x|}\bigg)\bigg|^{|\alpha|} 
\\
& = c_{n,\alpha}\frac{1}{|x|^{|\alpha|}}|(x_{1},x_{2},...x_{n})|^{|\alpha|}
\\
& = c_{n,\alpha}\frac{|x|^{|\alpha|}}{|x|^{|\alpha|}} 
= c_{n,\alpha}
\end{align}$$
So $|x^{\alpha}|\le c_{n,\alpha}|x|^{|\alpha|}$. 
Now we prove this inequality on the unit sphere.
$\vert x^{\alpha}\vert=|x_1^{\alpha_1}||x_2^{\alpha_{2}}|...|x_n^{\alpha_n}|\le\frac{1}{n}\sum_{k=1}^{n}|x_{k}|^{n\alpha_{k}}\le\frac{1}{n}\big(\sum_{k=1}^{n}|x_{k}|^{\alpha_{k}}\big)^{n}$
$\le\frac{1}{n}\big(\sum_{k=1}^{n}(1+|x_{k}|)^{\alpha_{k}}\big)^{n}\le\frac{1}{n}\big(\sum_{k=1}^{n}(1+|x|)^{\alpha_k}\big)^{n}\le\frac{1}{n}\big(n(1+|x|)^{|\alpha|})^{n}=n^{n-1}2^{n|\alpha|}$.
Note that I have used the Arithmetic Geometric Inequality to get the first inequality in the above proof. The second inequality follows from the inequality: $(x_{1}+x_{2}+...+x_{n})^{k}=\sum_{\alpha_{1}+...+\alpha_{n}=k}\binom{k}{\alpha_{1}, \alpha_{2},...,\alpha_{n}}x^{\alpha_{1}}...x^{\alpha_{n}}\ge x_{1}^{k}+...+x_{n}^{k}$ which is true when $x_{1},...x_{n}$ are all non-negative. Also note that $|x_{k}|\le1+|x_{k}|$ and $|x_{k}|\le \sqrt{x_{1}^{2}+...+x_{n}^{2}}=|x|$ for all $k=1,2,...,n$.
For the second inequality we attack as follows:
$|x|^{k}=\big(\sum_{k=1}^{n}|x_{k}|^{2}\big)^{\frac{k}{2}}\le\big(\sum_{k=1}^{n}|x_{k}|\big)^{\frac{2k}{2}}=\big(\sum_{k=1}^{n}|x_{k}|\big)^{k}=\sum_{|\beta|=k}\frac{k!}{\beta!}|x^{\beta}|\le\big(\sum_{|\beta|=k}\big(\frac{k!}{\beta!}\big)\big)\big(\sum_{|\beta|=k}|x^{\beta}|\big)=n^{k}\big(\sum_{|\beta|=k}|x^{\beta}|\big)$.
A: I think for the first inequality the constant is $1$ actually. I note $|\boldsymbol\alpha|=\alpha_1+\cdots+\alpha_n$ and $\mathbf{x}^{\boldsymbol\alpha}=x_1^{\alpha_1}\cdots x_n^{\alpha_n}$
The geometric mean inequality states that if $\beta_i\geq 0$ and $\sum_{i=1}^{n} \beta_i=1$ then for positive numbers $x_i\geq 0$ and $q>0$:
\begin{equation}
x_1^{\beta_1} \cdots x_d^{\beta_n} \leq (\sum_{i=1}^{n}\beta_i x_i^{q})^{1/q}
\end{equation}
Applying this for $|x_i|$, and $\beta_i=\alpha_i/|\boldsymbol\alpha|$ gives:
\begin{equation}
(|x_1|^{\alpha_1} \cdots |x_n|^{\alpha_n})^{1/|\boldsymbol\alpha|}\leq \frac{1}{|\boldsymbol\alpha|^{1/q}} (\sum_{i=1}^{n}\alpha_i |x_i|^{q})^{1/q}
\end{equation}
So:
\begin{equation}
|\mathbf{x}^{\boldsymbol\alpha}|\leq \frac{1}{|\boldsymbol\alpha|^{|\boldsymbol\alpha|/q}}(\sum_{i=1}^{d}\alpha_i |x_i|^{q})^{|\boldsymbol\alpha|/q}
\end{equation}
But $\alpha_i\leq |\boldsymbol\alpha|$ so:
\begin{equation}
\frac{1}{|\boldsymbol\alpha|^{|\boldsymbol\alpha|/q}}(\sum_{i=1}^{n}\alpha_i |x_i|^{q})^{|\boldsymbol\alpha|/q} \leq \frac{|\boldsymbol\alpha|^{|\boldsymbol\alpha|/q}}{|\boldsymbol\alpha|^{|\boldsymbol\alpha|/q}}(\sum_{i=1}^{n} |x_i|^{q})^{|\boldsymbol\alpha|/q} 
\end{equation}
Hence:
\begin{equation}
|\mathbf{x}^{\boldsymbol\alpha}|\leq (\sum_{i=1}^{n}|x_i|^{q})^{|\boldsymbol\alpha|/q}
\end{equation}
Taking $q=2$ gives:
\begin{equation}
|\mathbf{x}^{\boldsymbol\alpha}|\leq \|\mathbf{x}\|^{|\boldsymbol\alpha|}
\end{equation}
