Prove $d_1(x,y) \le \sqrt N \cdot d_2(x,y)$

Question

Let $d_1(x,y) = \sum_{k=1}^N|x(k)-y(k)|$ and $d_2=(\sum_{k=1}^N|x(k)-y(k)|^2)^{1/2}$ be metrics in $\mathbb{R}^N$. Prove that $d_1 \le \sqrt N \cdot d_2$

I said: $d_1(x,y) = \sqrt {(\sum_{k=1}^N|x(k)-y(k)|)^2} \le \sqrt{\sum_{k=1}^N|x(k)-y(k)|^2} = d_2(x,y) \le \sqrt N \cdot d_2(x,y)$ but it seems kind of odd. Can someone point out my mistake, if any, and point me in the right direction?

Answer 1

The first equality is wrong. Use Cauchy -Schwartz inequality: $\sum_{k=1}^{N} |a_k b_k| \leq (\sum_{k=1}^{N} |a_k|^{2})^{1/2} (\sum_{k=1}^{N} |b_k|^{2})^{1/2}$ with $a_k=1$ for all $k$ and $b_k=|x(k)-y(k)|$.