I am trying to show show that $\lvert x^{\beta}\rvert\le \lvert x \rvert^{\lvert\beta\rvert}$ where \beta is a multi-index. So I have to show $\lvert x_1^{\beta_1}\cdot\ldots\cdot x_n^{\beta_n}\rvert \le \sqrt{(x_1^2+\ldots+x_n^2)}^{\lvert \beta_1+\ldots+\beta_n\rvert}$. Should I show this by induction or is there a more straight-forward way?
I am stuck in the induction : $\lvert x_1^{\beta_1}\cdot\ldots\cdot x_n^{\beta_n}\rvert^2\lvert x_{n+1}^{\beta_{n+1}}\rvert^2 \le(x_1^2+\ldots +x_n^2)x_{n+1}^{2\beta_{n+1}}$
You can prove this directly, using the weighted AM-GM inequality and the Cauchy-Schwarz inequality.
Set $w_i = \frac{\beta_i}{|\beta|}$. Then $\sum w_i =1$ and therefore $$ |x_1|^{w_1} \cdots |x_n|^{w_n} \le w_1 |x_1| + \ldots + w_n |x_n| \\ \le \sqrt{w_1^2 + \ldots + w_n^2} \sqrt{x_1^2 + \ldots + x_n^2}\\ \le \sqrt{x_1^2 + \ldots + x_n^2}\, . $$ The last inequality holds because $$ \sqrt{w_1^2 + \ldots + w_n^2} \le \sqrt{w_1 + \ldots + w_n} = 1 \, . $$