Show that $\|x\|_\infty \leq \|x\|_2 \leq \|x\|_1$ for all $x \in \mathbb{R}^n$ and that equalities can occur for nonzero vectors.
Given \begin{align*} \|x\|_\infty &= \max_{1 \leq i \leq n} |x_i|,\\ \|x\|_2 &= \sqrt{\sum_{i=1}^nx_i^2},\\ \|x\|_1 &= \sum_{i=1}^n|x_i|. \end{align*} We first want to prove $\|x\|_\infty \leq \|x\|_2$. Without loss of generality, lets assume that the maximum is obtained for $i=1$. We then have $$\|x\|_\infty = \max_{1 \leq i \leq n} |x_i| = |x_1|= \sqrt{x_1^2}.$$ Now, for each $i = 1,2, \ldots, n$, we have $x_i^2 > 0$. Thus, $$\|x\|_\infty =|x_1| = \sqrt{x_1^2} \leq \sqrt{x_1^2 + \sum_{i=2}^nx_i^2} =\|x\|_2,$$ that is, $\|x\|_\infty \leq \|x\|_2$. Now, to prove $\|x\|_2 \leq \|x\|_1$, we define $$x^{(k)} = (0,0,\ldots,x_k,\ldots,0) \quad \forall k = 1,2, \ldots, n.$$ Then $$x = \sum_{k=1}^nx^{(k)},$$ and $$\|x\|_2 = \left\|\sum_{k=1}^nx^{(k)}\right\| \underbrace{\leq}_{\text{triangle inequality}} \sum_{k=1}^n||x^{(k)}||_2 = \sum_{k=1}^n\sqrt{\left(x^{(k)}\right)^2} = \sum_{k=1}^n|x^{(k)}| = \|x\|_1.$$ Finally, set $x = (x_1, 0,0 \ldots, 0) \in \mathbb{R}^n$ for $x_1 \neq 0$. Then \begin{align*} \|x\||_\infty &= \max_{1 \leq i \leq n} |x_i| = |x_1|,\\ \|x\|_2 &= \sqrt{x_1^2} = |x_1|,\\ \|x\|_1 &= |x_1|, \end{align*} therefore the equalities can occur for non-zero vectors.
Show that $\|x\|_1 \leq n \|x\|_\infty$ and $\|x\|_2 \leq \sqrt{n} \|x\|_\infty$ for $x \in \mathbb{R}^n$.
To see the first inequality, we have that for each $i = 1,2,\ldots,n$, the inequality $$|x_i| \leq \max_{1 \leq i \leq n} |x_i|,$$ holds. This means that $$\sum_{k = 1}^n|x_i| \leq n \max_{1 \leq i \leq n} |x_i|,$$ and it follows that $\|x\|_1 \leq n\|x\|_\infty$. To see the other inequality, we have $$\max_{1 \leq i \leq n} |x_i| = \max_{1 \leq i \leq n} \sqrt{x_i^2}.$$ Now, on the other hand, $$\|x\|_2^2 = \sum_{i=1}^n x_i^2 \text{ and } \|x\|_\infty^2 = \max_{1 \leq i \leq n} x_i^2.$$ By a similar reasoning, we have that for each $i = 1,2,\ldots,n$, the inequality $$x_i^2 \leq \max_{1 \leq i \leq n} x_i^2,$$ holds. Therefore, $$\sum_{i=1}^n x_i^2 \leq n \max_{1 \leq i \leq n} x_i^2.$$ We now take the square roots to get $$\sqrt{\sum_{i=1}^n x_i^2} \leq \sqrt{n} \max_{1 \leq i \leq n} \sqrt{x_i^2},$$ that is, $\|x\|_2 \leq \sqrt{n} \|x\|_\infty$. Is this the right reasoning?
Thanks in advance.