[proofplan] Fix two points in $X\times Y$ and reduce the comparison to an inequality for two nonnegative [real numbers](/page/Real%20Numbers). The two relevant numbers are the coordinate distances $d_X(x_1,x_2)$ and $d_Y(y_1,y_2)$. The lower bound follows because the larger coordinate distance has $p$-th power bounded by the sum of the two $p$-th powers, while the upper bound follows because both coordinate distances are bounded by their maximum. [/proofplan]

[step:Reduce the product distances to two nonnegative real numbers] Fix $(x_1,y_1),(x_2,y_2)\in X\times Y$. Define \begin{align*} a := d_X(x_1,x_2) \end{align*} and \begin{align*} b := d_Y(y_1,y_2). \end{align*} Since $d_X$ and $d_Y$ are metrics, $a,b\in [0,\infty)$. Define \begin{align*} m := \max\{a,b\}. \end{align*} By the definitions of $d_p$ and $d_\infty$, \begin{align*} d_p((x_1,y_1),(x_2,y_2)) = (a^p+b^p)^{1/p} \end{align*} and \begin{align*} d_\infty((x_1,y_1),(x_2,y_2)) = m. \end{align*} [/step]

[step:Bound the maximum by the $p$-product distance] Since $m=\max\{a,b\}$, either $m=a$ or $m=b$. In both cases, \begin{align*} m^p \le a^p+b^p. \end{align*} The function $t\mapsto t^{1/p}$ is increasing on $[0,\infty)$, so taking $p$-th roots gives \begin{align*} m \le (a^p+b^p)^{1/p}. \end{align*} Using the identities from the previous step, this is exactly \begin{align*} d_\infty((x_1,y_1),(x_2,y_2)) \le d_p((x_1,y_1),(x_2,y_2)). \end{align*} [/step]

[step:Bound the $p$-product distance by the maximum]Since $a\le m$ and $b\le m$, and since $p\ge 1$, we have \begin{align*} a^p \le m^p \end{align*} and \begin{align*} b^p \le m^p. \end{align*} Adding these inequalities gives \begin{align*} a^p+b^p \le 2m^p. \end{align*} Taking $p$-th roots, again using monotonicity of $t\mapsto t^{1/p}$ on $[0,\infty)$, yields \begin{align*} (a^p+b^p)^{1/p} \le (2m^p)^{1/p}=2^{1/p}m. \end{align*} Substituting the definitions of $d_p$ and $d_\infty$ gives \begin{align*} d_p((x_1,y_1),(x_2,y_2)) \le 2^{1/p}d_\infty((x_1,y_1),(x_2,y_2)). \end{align*}[/step]

[guided]With $a=d_X(x_1,x_2)$, $b=d_Y(y_1,y_2)$, and $m=\max\{a,b\}$ as above, the goal is to compare the sum $a^p+b^p$ with $m^p$. The point of introducing $m$ is that it is a single number controlling both coordinate distances. By definition of maximum, \begin{align*} a\le m \end{align*} and \begin{align*} b\le m. \end{align*} Because $p\ge 1$ and all three numbers $a,b,m$ are nonnegative, raising to the $p$-th power preserves the inequalities: \begin{align*} a^p\le m^p \end{align*} and \begin{align*} b^p\le m^p. \end{align*} Adding these two inequalities gives \begin{align*} a^p+b^p\le m^p+m^p=2m^p. \end{align*} Now we pass from $p$-th powers back to distances. The map $t\mapsto t^{1/p}$ is increasing on $[0,\infty)$, so taking $p$-th roots preserves the inequality: \begin{align*} (a^p+b^p)^{1/p}\le (2m^p)^{1/p}. \end{align*} Since $m\ge 0$, the right-hand side simplifies as \begin{align*} (2m^p)^{1/p}=2^{1/p}m. \end{align*} Using \begin{align*} d_p((x_1,y_1),(x_2,y_2))=(a^p+b^p)^{1/p} \end{align*} and \begin{align*} d_\infty((x_1,y_1),(x_2,y_2))=m, \end{align*} we obtain \begin{align*} d_p((x_1,y_1),(x_2,y_2))\le 2^{1/p}d_\infty((x_1,y_1),(x_2,y_2)). \end{align*}[/guided]

[step:Combine the two inequalities] The lower bound and upper bound obtained above hold for the fixed pair $(x_1,y_1),(x_2,y_2)\in X\times Y$. Since the pair was arbitrary, for every $(x_1,y_1),(x_2,y_2)\in X\times Y$, \begin{align*} d_\infty((x_1,y_1),(x_2,y_2)) \le d_p((x_1,y_1),(x_2,y_2)) \le 2^{1/p}d_\infty((x_1,y_1),(x_2,y_2)). \end{align*} This proves the comparison of the two product metrics. [/step]