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 $\|x\|_k = (\sum_{i}|x_i|^k)^\frac{1}{k}$.

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*}
