Give examples of vectors in $\mathbb R^n$ for which equality holds in parts $(a)$ and $(b)$. Let $x \in\mathbb R^n$ and show that 
$(a)$ $||x||_1 \le n||x||_\infty$
$(b)\;||x||_2 \le \sqrt{n}||x||_\infty $
Proof
Suppose $$\lVert \widehat x\rVert_1= \sum^{n}_{i=1} |x_i| \Rightarrow \lVert (-2,-3)\rVert_1=|-2|+|3|=5 \\\text{Let}\\ \lVert \widehat x\rVert_\infty= \text{max}|x_i|=3 \Rightarrow n \cdot3=2\cdot3=6 \\ \Rightarrow \\|x_i| \le \text{max}|x_i|\\\text{Thus}\\|x_1|+|x_2|+|x_i|+|x_n|\le |x_i| \text{with} |x_i| \text{maximum} \qquad\square$$
Is there any way to streamline this process? Is there any other examples one can use to solve this question?
 A: You can easily show that the inequality is true for all vectors in $\mathbb{R}^n$.
\begin{align*}
\Vert x\Vert_1 = \sum_{i=1}^{n}|x_i| ≤ \sum_{i=1}^{n}\max\{|x_1|,|x_2|,..., |x_n|\} = n\Vert x \Vert_\infty
\end{align*}
\begin{align*}
\Vert x\Vert_2 = \sqrt{\sum_{i=1}^{n}x_i^2} ≤ \sqrt{\sum_{i=1}^{n}\max\{|x_1|,|x_2|,..., |x_n|\}^2} = \sqrt{n\Vert x \Vert_{\infty}^2} = \sqrt{n}\Vert x \Vert_\infty
\end{align*}
On the other hand, equality is more interesting. Both equalities hold precisely when the absolute values of all components of the vector are equal. The key idea is that $\Vert \cdot \Vert_\infty$ depends only on the largest component, but $\Vert \cdot \Vert_n$ for $n \in \mathbb{N}$ depends on all components. This means as the "other" components get greater in size relative to the largest component, $\Vert \cdot \Vert_\infty$ won't change but $\Vert \cdot \Vert_n$ will get larger.
$\Rightarrow$ Let $x \in \mathbb{R}^n.\ $Suppose $\Vert x \Vert_1 = n\Vert x \Vert_\infty$. Then $\sum_{i=1}^n |x_i| = n\max\{|x_i|, 1≤i≤n\}$. Suppose with a view to contradiction that $|x_p| \neq |x_q|$. Without loss of generality, suppose $|x_p| > |x_q|$, and $\max\{|x_i|, 1≤i≤n\} = |x_p|$. Now $\Vert x \Vert _1 = \sum_{i=1}^n |x_i| ≤ (n-1)|x_p|+|x_q|< n|x_p| = n\max\{|x_i|, 1≤i≤n\} = n\Vert x \Vert_\infty$, a contradiction.
$\Leftarrow$ Let $x\in \mathbb{R}^n$. Suppose the absolute values of all components of x are equal. The proof is trivial and left as an exercise to the reader.
Similar proofs are used for the second inequality.
Have fun proving them :)
