Showing $||x||_{\infty} \leq ||x||_{2} \leq ||x||_{1}$ I am currently reworking the following exercise that was worked during my previous class's lecture on normed vector spaces:

Show that $||x||_{\infty} \leq ||x||_{2} \leq ||x||_{1}$ for any $x \in \mathbb{R}^n$.

After class, I asked if this is the same as proving that $||a||_p\le ||a||_1$ for any $p \leq 1$ or more generally, $||a||_q\le ||a||_p$ whenever $p \leq q$ . My professor then asked me to finish the following computation and report back my findings:
$(\sum_{i=0}^n |a_i|^p)^{1/p}\le (\sum_{i=0}^{n-1} |a_i|^p)^{1/p} +(|a_n|^p)^{1/p}\\\le (\sum_{i=0}^{n-2} |a_i|^p)^{1/p} +(|a_{n-1}|^p)^{1/p} + (|a_n|^p)^{1/p} \\ ...\\...\\$
However, as I have made progress I have a few questions.
$\bullet$ Is this using Minkowski's inequality repeatedly?
$\bullet$ I don't fully understand the step  $(\sum_{i=0}^n |a_i|^p)^{1/p}\le (\sum_{i=0}^{n-1} |a_i|^p)^{1/p} +(|a_n|^p)^{1/p}$. Why are we adding the $p$ norm of the $n^{th}$ term to ($\sum_{i=0}^{n-1} |a_i|^p)^{1/p}$?
 A: Note that for $a, b> 0$ we have $(a + b)^{1/x} \le a^{1/x} + b^{1/x}$ for $x \ge 1$. Hence you have $\left(\sum|a_i|^p\right)^{1/p} =  \|a \|_{p} \le \| a\|_1 = \sum |a_i|$.
A: For the first question the answer is simply yes.
For the second we have that Minkowski's inequality for finite sums states that
$$\left(\sum_{k=0}^{n}|\xi_{k}+\eta_{k}|^{p}\right)^{\frac{1}{p}}\leq\left(\sum_{k=0}^{n}|\xi_{k}|^{p}\right)^{\frac{1}{p}}+\left(\sum_{k=0}^{n}|\eta_{k}|^{p}\right)^{\frac{1}{p}} \tag{1}$$
Now starting from the left hand side of your expression we could write $\xi_{k}=a_{k}$ for $0\leq k \leq n$ and $\eta_{k}=0$ for all such $k$, then we would have your expression on the left hand side of (1), but, Minkowski's inequality would not tell us much. However, by changing the last element: $\xi_{n}=0$ and $\eta_{n}=a_{n}$ we have not changed the left hand side of (1) but we have changed the right. In fact we have
$$\left(\sum_{k=0}^{n}|a_{k}|^{p}\right)^{\frac{1}{p}}\leq\left(\sum_{k=0}^{n-1}|a_{k}|^{p}\right)^{\frac{1}{p}}+(|a_{n}|^p)^\frac{1}{p}=\left(\sum_{k=0}^{n-1}|a_{k}|^{p}\right)^{\frac{1}{p}}+|a_{n}|$$
Notice here that
$$\left(\sum_{k=0}^{n}|\eta_{k}|^{p}\right)^{\frac{1}{p}}=(|a_n|^{p})^\frac{1}{p}$$
because we have $\eta_{k}=0$ for all $0\leq k < n$ and $\eta_{n}=a_{n}$. We have also changed the index of the sum containing the "$a$'s" from $n$ to $n-1$ because the last element there was $0$.
