$L^p$ norm equivalence I am trying to show $A = ||Du||_{L^p}$ is equivalent to $B = \sum_{|\alpha| = 1} ||D^{\alpha} u ||_{L^p}$ in the sense of norms where $ p \geq 1$ by using Jensen's inequality.
$B^p = \sum_{i=1}^n \left( \int_U |u_{x_i}|^p \right)^{1/p} = \left[ \left(\int_U |u_{x_1}|^p\right)^{1/p} + \cdots + \left(\int_U |u_{x_n}|^p\right)^{1/p}\right]= ( a_1 + \cdots + a_n)^p $
and it would be nice if I can somehow show $ A^p = \int_U |Du|^p = \int_U \left(\sqrt{u_{x_1}^2 + \cdots + u_{x_n}^2}\right) ^p$ is greater than $n^{p-1} (a_1^p + \cdots + a_n^p)$ in order to use Jensen's inequality
$$ (a_1^p + \cdots a_n^p) \leq (a_1+\cdots a_n)^p \leq  n^{p-1} (a_1^p + \cdots a_n^p)$$
Any tips please?
 A: Assume $2\le p$ (a reverse argument holds for $p\le 2$). By the arithmetic mean inequalities, $$\frac{|a_1|+\cdots+|a_n|}{n}\le\sqrt{\frac{|a_1|^2+\cdots+|a_n|^2}{n}}\le\sqrt[p]{\frac{|a_1|^p+\cdots+|a_n|^p}{n}}\le\max_i|a_i|\tag{1}$$ Hence
$$|a_1+\cdots+a_n|\le n^{\frac{1}{2}}\sqrt{|a_1|^2+\cdots+|a_n|^2}\le n^{1-\frac{1}{p}}\sqrt[p]{|a_1|^p+\cdots+|a_n|^p}\tag{2} $$
$$\sqrt[p]{|a_1|^p+\cdots+|a_n|^p}\le (|a_1|+\cdots+|a_n|)\le n^{\frac{1}{2}}\sqrt{|a_1|^2+\cdots+|a_n|^2}\tag{3}$$
Therefore
\begin{align}A=\|Du\|_p&=\sqrt[p]{\int(|u_1|^2+\cdots+|u_n|^2)^{p/2}}\\&\le n^{\frac{1}{2}-\frac{1}{p}}\sqrt[p]{\int(|u_1|^p+\cdots+|u_n|^p)}\qquad\textrm{by (2)}\\&\le n^{\frac{1}{2}}\max_i\|u_i\|_p\le n^{\frac{1}{2}}\sum_i\|u_i\|_p=n^{\frac{1}{2}}B\qquad\textrm{by (1)}\end{align}
Conversely, \begin{align}B=\sum_i\|u_i\|_p&\le n^{1-\frac{1}{p}}\sqrt[p]{\|u_1\|_p^p+\cdots+\|u_n\|_p^p}\qquad\textrm{by (2)}\\
&=n^{1-\frac{1}{p}}\sqrt[p]{\int|u_1|^p+\cdots|u_n|^p}\\
&\le  n^{\frac{3}{2}-\frac{1}{p}}\sqrt[p]{\int\left(|u_1|^2+\cdots+|u_n|^2\right)^{p/2}}\qquad \textrm{by (3)}\\
&=n^{\frac{3}{2}-\frac{1}{p}}\|Du\|_p=n^{\frac{3}{2}-\frac{1}{p}}A\end{align}
