Given two points $x,y\in\textbf{R}^{n}$, prove the following inequalities \begin{align*} d_{2}(x,y) \leq d_{1}(x,y) \leq \sqrt{n}d_{2}(x,y) \end{align*}
MY ATTEMPT
According to its corresponding definitions, we have
\begin{align*} \begin{cases} d_{1}(x,y) = |x_{1} - y_{1}| + |x_{2} - y_{2}| + \ldots + |x_{n} - y_{n}|\\\\ d_{2}(x,y) = (|x_{1} - y_{1}|^{2} + |x_{2} - y_{2}|^{2} + \ldots + |x_{n} - y_{n}|^{2})^{1/2} \end{cases} \end{align*}
Consequently, one has that \begin{align*} d^{2}_{1}(x,y) - d^{2}_{2}(x,y) = \sum_{i = 1}^{n}|x_{i} - y_{i}|\sum_{j\neq i}|x_{j} - y_{j}| \geq 0 \Rightarrow d_{1}(x,y)\geq d_{2}(x,y) \end{align*}
Similarly, we have that \begin{align*} nd^{2}_{2}(x,y) - d^{2}_{1}(x,y) = \sum_{k=1}^{n}(n-1)|x_{k} - y_{k}|^{2} - \sum_{i=1}^{n}|x_{i}-y_{i}|\sum_{j\neq i}|x_{j} - y_{j}| \end{align*} then I get stuck.
Can someone please help me finishing this exercise?
$\newcommand{\bb}[1]{\left( #1 \right)}$ For the right inequality, we can apply Cauchy-Schwarz inequality: $$ d_1(x,y) = \sum_{i=1}^n |x_i - y_i| \leq \bb{\sum_{i=1}^n |x_i - y_i|^2}^\frac{1}{2}\bb{\sum_{i=1}^n 1}^\frac{1}{2} = \sqrt{n}d_2(x,y) $$