We show that $\displaystyle d_r \left({x, y}\right) \ge d_t \left({x, y}\right)$, which is equivalent to proving that: :$\displaystyle \left({\sum_{i \mathop = 1}^n \left|{x_i - y_i}\right|^r}\right)^{1/r} \ge \left({\sum_{i \mathop = 1}^n \left|{x_i - y_i}\right|^t}\right)^{1/t}$ Let $\forall i \in \left[{1 \,.\,.\, n}\right]: s_i = \left|{x_i - y_i}\right|$. Suppose $s_k = 0$ for some $k \in \left[{1 \,.\,.\, n}\right]$. {{improve|Putting off the analysis of s_k being 0 until near the end of the proof should shorten it without losing any clarity.}} Then the problem reduces to the equivalent one of showing that: : $\displaystyle \left({\sum_{i \mathop = 1}^{n-1} \left|{x_i - y_i}\right|^r}\right)^{1/r} \ge \left({\sum_{i \mathop = 1}^{n-1} \left|{x_i - y_i}\right|^t }\right)^{1/t}$ that is, of reducing the index by $1$. Note that when $n = 1$, from simple algebra $d_r \left({x, y}\right) = d_t \left({x, y}\right)$. So, let us start with the assumption that $\forall i \in \left[{1 \,.\,.\, n}\right]: s_i > 0$. Let $\displaystyle u = \sum_{i \mathop = 1}^n \left|{x_i - y_i}\right|^r = \sum_{i \mathop = 1}^n s_i^r$, and $v = \dfrac 1 r$. From Derivative of Function to Power of Function, $D_r \left({u^v}\right) = v u^{v-1} D_r \left({u}\right) + u^v \ln u D_r  \left({v}\right)$. Here: :$\displaystyle D_r \left({u}\right) = \sum_{i \mathop = 1}^n s_i^r \ln s_i$ from Derivative of Exponential Function and Sum Rule for Derivatives :$D_r \left({v}\right) = - \dfrac 1 {r^2}$ from Power Rule for Derivatives In the case where $r=1$, we have: :$D_r \left({u^v}\right) = 0$ When $r > 1$, we have: {{begin-eqn}} {{eqn | l = D_r \left({\left({\sum_{i \mathop = 1}^n s_i^r}\right)^{1/r} }\right)       | r = \dfrac 1 r \left({\sum_{i \mathop = 1}^n s_i^r}\right)^{1/ \left({r - 1}\right) } \left({\sum_{i \mathop = 1}^n s_i^r \ln s_i}\right) - \dfrac {\left({\sum_{i \mathop = 1}^n s_i^r}\right)^{1/r} \ln \left({\sum_{i \mathop = 1}^n s_i^r}\right)} {r^2}       | c =  }} {{eqn | r = \dfrac {\left({\sum_{i \mathop = 1}^n s_i^r}\right)^{1/r} } r \left({\dfrac {\sum_{i \mathop = 1}^n s_i^r \ln s_i} {\sum_{i \mathop = 1}^n s_i^r} - \dfrac {\ln \left({\sum_{i \mathop = 1}^n s_i^r}\right)} r}\right)       | c =  }} {{eqn | r = \dfrac {\left({\sum_{i \mathop = 1}^n s_i^r}\right)^{1/r} } r \left({\dfrac {r \left({\sum_{i \mathop = 1}^n s_i^r \ln s_i}\right) - \left({\sum_{i \mathop = 1}^n s_i^r}\right) \ln \left({\sum_{i \mathop = 1}^n s_i^r}\right)} {r \left({\sum_{i \mathop = 1}^n s_i^r}\right)} }\right)       | c =  }} {{eqn | r = K \left({r \left({\sum_{i \mathop = 1}^n s_i^r \ln s_i}\right) - \left({\sum_{i \mathop = 1}^n s_i^r}\right) \ln \left({\sum_{i \mathop = 1}^n s_i^r}\right)}\right)       | c = where $\displaystyle K = \dfrac {\left({\sum_{i \mathop = 1}^n s_i^r}\right)^{1/r} } {r^2 \left({\sum_{i \mathop = 1}^n s_i^r}\right)} > 0$ }} {{eqn | r = K \left({\sum_{i \mathop = 1}^n s_i^r \ln \left({s_i^r}\right) - \left({\sum_{i \mathop = 1}^n s_i^r}\right) \ln \left({\sum_{i \mathop = 1}^n s_i^r}\right)}\right)       | c = Logarithms of Powers }} {{eqn | r = K \left({\sum_{j \mathop = 1}^n \left({s_j^r \left({\ln \left({s_j^r}\right) - \ln \left({\sum_{i \mathop = 1}^n s_i^r}\right)}\right)}\right)}\right)       | c =  }} {{eqn | r = K \left({\sum_{j \mathop = 1}^n \left({s_j^r \ln \left({\frac {s_j^r} {\sum_{i \mathop = 1}^n s_i^r} }\right)}\right)}\right)       | c =  }} {{end-eqn}} where $K > 0$ because all of $s_i, r > 0$. For the same reason, $\displaystyle \dfrac{s_j^r} {\sum_{i \mathop = 1}^n s_i^r} < 1$ for all $j \in \left\{ {1, \ldots, n}\right\}$. From Logarithm of 1 is 0 and Logarithm is Strictly Increasing, their logarithms are therefore negative. So for $r > 1$: : $\displaystyle D_r \left({\left({\sum_{i \mathop = 1}^n s_i^r}\right)^{1/r}}\right) < 0$ So, from Derivative of Monotone Function, it follows that (given the conditions on $r$ and $s_i$) $\displaystyle \left({\sum_{i \mathop = 1}^n s_i^r}\right)^{1/r}$ is decreasing. As we assumed $r \le t$, we have $d_r \left({x, y}\right) \ge d_t \left({x, y}\right)$. {{qed|lemma}}