How to prove $n^{\frac{1}{q}}\|x\|_p \le n^{\frac{1}{p}} \|x\|_q$ if $1\le p\le q$? Let $1\le p\le q$ and $x\in \mathbb{R}^n$. Show that
$$
  n^{\frac{1}{q}}\|x\|_p \le n^{\frac{1}{p}} \|x\|_q,
$$
where $\|x\|_k = (\sum_{i}|x_i|^k)^\frac{1}{k}$.
 A: That's a standard application of Hölder's inequality. Let $\alpha := \frac qp \ge 1$ and $\beta = \frac 1{1 - \frac 1\alpha} = \frac q{q-p}$
\begin{align*}
  \def\norm#1{\left\|#1\right\|}\def\abs#1{\left|#1\right|}
  \norm{x}_p &= \left(\sum_i \abs{x_i}^p\right)^{1/p}\\
             &= \left(\sum_i \abs{x_i}^p \cdot 1 \right)^{1/p}\\
             &\le \left(\norm{(\abs{x_i}^p)_i}_\alpha \norm{1}_\beta\right)^{1/p}\\
             &= \left(\sum_i \abs{x_i}^{p\alpha}\right)^{1/p\alpha}
                  \cdot \left(\sum_i 1^{\beta}\right)^{1/p\beta}\\
             &= \norm x_q \cdot n^{(q-p)/pq}\\
             &= \norm x_q \cdot  n^{1/p - 1/q}
\end{align*}
A: Here is a proof using Jensen's inequality. Note that $\phi(x) = x^{\tfrac q p}$ is convex since $q \ge p$. Now we can apply Jensen's ineq $\phi(E X) \le E\phi(X)$. Our distribution is uniform over the set $\{\lvert x_i\rvert^p \mid  i=1,\dots, n\}$.
$$\begin{align}
 & \left[ \frac{1}{n}\sum_{i=1}^{n} \lvert x_i\rvert^p  \right]^{q/p} &\le  \frac{1}{n}\sum_{i=1}^{n} \lvert x_i\rvert^q \\
& \implies \frac{1}{n^{1/p}} \lVert x\rVert_p &\le \frac{1}{n^{1/q}} \lVert x\rVert_q
\end{align}$$
