[proofplan] We verify the metric axioms on $X\times Y$ coordinatewise. Nonnegativity, symmetry, and identity of indiscernibles follow directly from the corresponding axioms for $d_X$ and $d_Y$. The only substantive point is the triangle inequality: for $1\le p<\infty$ we prove the required two-coordinate Minkowski estimate, and for $p=\infty$ we use the elementary inequality $\max\{a+c,b+d\}\le \max\{a,b\}+\max\{c,d\}$ for nonnegative [real numbers](/page/Real%20Numbers). [/proofplan]

[step:Verify nonnegativity and symmetry from the coordinate metrics] Let $(x_1,y_1),(x_2,y_2)\in X\times Y$. First suppose $1\le p<\infty$. Since $d_X(x_1,x_2)\ge 0$ and $d_Y(y_1,y_2)\ge 0$, the quantity \begin{align*} d_X(x_1,x_2)^p+d_Y(y_1,y_2)^p \end{align*} is nonnegative, so $d_p((x_1,y_1),(x_2,y_2))\ge 0$. Since $d_X$ and $d_Y$ are symmetric, \begin{align*} d_X(x_1,x_2)=d_X(x_2,x_1) \end{align*} and \begin{align*} d_Y(y_1,y_2)=d_Y(y_2,y_1). \end{align*} Substituting these equalities into the finite-$p$ formula gives \begin{align*} d_p((x_1,y_1),(x_2,y_2))=d_p((x_2,y_2),(x_1,y_1)). \end{align*} Now suppose $p=\infty$. The maximum of two nonnegative real numbers is nonnegative, so $d_\infty((x_1,y_1),(x_2,y_2))\ge 0$. The same symmetry equalities for $d_X$ and $d_Y$ give \begin{align*} d_\infty((x_1,y_1),(x_2,y_2))=d_\infty((x_2,y_2),(x_1,y_1)). \end{align*} [/step]

[step:Check identity of indiscernibles coordinate by coordinate] Let $(x_1,y_1),(x_2,y_2)\in X\times Y$. If $(x_1,y_1)=(x_2,y_2)$, then $x_1=x_2$ and $y_1=y_2$. Since $d_X$ and $d_Y$ are metrics, \begin{align*} d_X(x_1,x_2)=0 \end{align*} and \begin{align*} d_Y(y_1,y_2)=0. \end{align*} Therefore $d_p((x_1,y_1),(x_2,y_2))=0$ for every $1\le p\le\infty$. Conversely, first suppose $1\le p<\infty$ and \begin{align*} d_p((x_1,y_1),(x_2,y_2))=0. \end{align*} By the finite-$p$ formula, \begin{align*} d_X(x_1,x_2)^p+d_Y(y_1,y_2)^p=0. \end{align*} Both summands are nonnegative, so each summand is zero: \begin{align*} d_X(x_1,x_2)^p=0 \end{align*} and \begin{align*} d_Y(y_1,y_2)^p=0. \end{align*} Since $p\ge 1$, this implies \begin{align*} d_X(x_1,x_2)=0 \end{align*} and \begin{align*} d_Y(y_1,y_2)=0. \end{align*} The identity of indiscernibles for $d_X$ and $d_Y$ gives $x_1=x_2$ and $y_1=y_2$, hence $(x_1,y_1)=(x_2,y_2)$. Now suppose $p=\infty$ and \begin{align*} d_\infty((x_1,y_1),(x_2,y_2))=0. \end{align*} Then \begin{align*} \max\{d_X(x_1,x_2),d_Y(y_1,y_2)\}=0. \end{align*} Since both entries in the maximum are nonnegative, both are zero. Thus $d_X(x_1,x_2)=0$ and $d_Y(y_1,y_2)=0$, so $x_1=x_2$ and $y_1=y_2$. Hence $(x_1,y_1)=(x_2,y_2)$. [/step]

[step:Prove the two-coordinate Minkowski estimate for finite $p$]We prove the following estimate: for every $1\le p<\infty$ and every $a,b,c,d\in[0,\infty)$, \begin{align*} \left((a+c)^p+(b+d)^p\right)^{1/p}\le \left(a^p+b^p\right)^{1/p}+\left(c^p+d^p\right)^{1/p}. \end{align*} For $p=1$, the estimate is equality: \begin{align*} (a+c)+(b+d)=(a+b)+(c+d). \end{align*} Assume now that $1<p<\infty$. Define $q\in(1,\infty)$ by \begin{align*} q=\frac{p}{p-1}. \end{align*} Then \begin{align*} \frac{1}{p}+\frac{1}{q}=1. \end{align*} We first use the two-term Hölder inequality: for all $r,s,u,v\in[0,\infty)$, \begin{align*} ru+sv\le \left(r^p+s^p\right)^{1/p}\left(u^q+v^q\right)^{1/q}. \end{align*} For completeness, this follows from [Young's inequality](/theorems/244) $AB\le A^p/p+B^q/q$ for $A,B\ge 0$, which is obtained by minimizing the function $\phi_A:(0,\infty)\to\mathbb R$ given by $\phi_A(B)=A^p/p+B^q/q-AB$ and using $\phi_A'(B)=B^{q-1}-A$. Normalizing by \begin{align*} R=\left(r^p+s^p\right)^{1/p} \end{align*} and \begin{align*} U=\left(u^q+v^q\right)^{1/q} \end{align*} when $R,U>0$, and treating the cases $R=0$ or $U=0$ separately, Young's inequality gives the displayed Hölder inequality. Define \begin{align*} S=\left((a+c)^p+(b+d)^p\right)^{1/p}. \end{align*} If $S=0$, the desired estimate follows because the right-hand side is nonnegative. Assume $S>0$. Expanding one power of each summand gives \begin{align*} S^p=a(a+c)^{p-1}+c(a+c)^{p-1}+b(b+d)^{p-1}+d(b+d)^{p-1}. \end{align*} Grouping the terms with coefficients $a,b$ and the terms with coefficients $c,d$, then applying the two-term Hölder inequality twice, gives \begin{align*} S^p\le \left(a^p+b^p\right)^{1/p}\left((a+c)^{(p-1)q}+(b+d)^{(p-1)q}\right)^{1/q}+\left(c^p+d^p\right)^{1/p}\left((a+c)^{(p-1)q}+(b+d)^{(p-1)q}\right)^{1/q}. \end{align*} Since $(p-1)q=p$, the common second factor is \begin{align*} \left((a+c)^p+(b+d)^p\right)^{1/q}=S^{p/q}=S^{p-1}. \end{align*} Thus \begin{align*} S^p\le \left(\left(a^p+b^p\right)^{1/p}+\left(c^p+d^p\right)^{1/p}\right)S^{p-1}. \end{align*} Dividing by $S^{p-1}>0$ proves the two-coordinate Minkowski estimate.[/step]

[guided]We need an inequality that says the $p$-length of a sum of two nonnegative coordinate vectors is at most the sum of their $p$-lengths. Written in coordinates, this is exactly the estimate \begin{align*} \left((a+c)^p+(b+d)^p\right)^{1/p}\le \left(a^p+b^p\right)^{1/p}+\left(c^p+d^p\right)^{1/p}. \end{align*} The case $p=1$ contains no hidden inequality: both sides reduce to the same sum, \begin{align*} (a+c)+(b+d)=(a+b)+(c+d). \end{align*} Now assume $1<p<\infty$. Define the conjugate exponent $q\in(1,\infty)$ by \begin{align*} q=\frac{p}{p-1}. \end{align*} Then \begin{align*} \frac{1}{p}+\frac{1}{q}=1. \end{align*} The estimate rests on the two-term Hölder inequality. We record it in the exact form needed here: for $r,s,u,v\in[0,\infty)$, \begin{align*} ru+sv\le \left(r^p+s^p\right)^{1/p}\left(u^q+v^q\right)^{1/q}. \end{align*} Here is the verification. Young's inequality says that for $A,B\ge 0$, \begin{align*} AB\le \frac{A^p}{p}+\frac{B^q}{q}. \end{align*} To justify it, fix $A\ge 0$ and consider the function $\phi_A:(0,\infty)\to\mathbb R$ defined by \begin{align*} \phi_A(B)=\frac{A^p}{p}+\frac{B^q}{q}-AB. \end{align*} Its derivative is \begin{align*} \phi_A'(B)=B^{q-1}-A. \end{align*} The minimum occurs at $B=A^{1/(q-1)}$, equivalently at $B^q=A^p$, and the minimum value is $0$ because $1/p+1/q=1$. The endpoint case $B=0$ is immediate, so Young's inequality holds for all $A,B\ge 0$. To derive Hölder, define \begin{align*} R=\left(r^p+s^p\right)^{1/p} \end{align*} and \begin{align*} U=\left(u^q+v^q\right)^{1/q}. \end{align*} If $R=0$ or $U=0$, then the products $ru$ and $sv$ are both zero, and the Hölder inequality follows. If $R>0$ and $U>0$, apply Young's inequality to $A=r/R$, $B=u/U$ and again to $A=s/R$, $B=v/U$. Adding the two inequalities gives \begin{align*} \frac{ru+sv}{RU}\le \frac{r^p+s^p}{pR^p}+\frac{u^q+v^q}{qU^q}. \end{align*} By the definitions of $R$ and $U$, the right-hand side is \begin{align*} \frac{1}{p}+\frac{1}{q}=1. \end{align*} Multiplying by $RU$ gives the two-term Hölder inequality. Now define \begin{align*} S=\left((a+c)^p+(b+d)^p\right)^{1/p}. \end{align*} If $S=0$, then the left-hand side of the desired inequality is zero, while the right-hand side is nonnegative. Hence the estimate holds in that case. Assume $S>0$. The useful move is to write $S^p$ with one copy of each coordinate separated: \begin{align*} S^p=(a+c)(a+c)^{p-1}+(b+d)(b+d)^{p-1}. \end{align*} Expanding the two sums gives \begin{align*} S^p=a(a+c)^{p-1}+c(a+c)^{p-1}+b(b+d)^{p-1}+d(b+d)^{p-1}. \end{align*} Group the terms involving $a,b$ together and the terms involving $c,d$ together: \begin{align*} S^p=\left(a(a+c)^{p-1}+b(b+d)^{p-1}\right)+\left(c(a+c)^{p-1}+d(b+d)^{p-1}\right). \end{align*} Apply the two-term Hölder inequality to the first parenthesis with \begin{align*} (r,s,u,v)=(a,b,(a+c)^{p-1},(b+d)^{p-1}) \end{align*} and to the second parenthesis with \begin{align*} (r,s,u,v)=(c,d,(a+c)^{p-1},(b+d)^{p-1}). \end{align*} This yields \begin{align*} S^p\le \left(a^p+b^p\right)^{1/p}\left((a+c)^{(p-1)q}+(b+d)^{(p-1)q}\right)^{1/q}+\left(c^p+d^p\right)^{1/p}\left((a+c)^{(p-1)q}+(b+d)^{(p-1)q}\right)^{1/q}. \end{align*} Because $q=p/(p-1)$, we have $(p-1)q=p$. Therefore the shared second factor is \begin{align*} \left((a+c)^p+(b+d)^p\right)^{1/q}=S^{p/q}=S^{p-1}. \end{align*} Substituting this into the previous inequality gives \begin{align*} S^p\le \left(\left(a^p+b^p\right)^{1/p}+\left(c^p+d^p\right)^{1/p}\right)S^{p-1}. \end{align*} Since $S>0$, division by $S^{p-1}$ is valid and gives \begin{align*} S\le \left(a^p+b^p\right)^{1/p}+\left(c^p+d^p\right)^{1/p}. \end{align*} This is the desired two-coordinate Minkowski estimate.[/guided]

[step:Apply the finite Minkowski estimate to prove the finite $p$ triangle inequality] Assume $1\le p<\infty$. Let $(x_1,y_1),(x_2,y_2),(x_3,y_3)\in X\times Y$. Define four nonnegative real numbers \begin{align*} a=d_X(x_1,x_2) \end{align*} \begin{align*} b=d_Y(y_1,y_2) \end{align*} \begin{align*} c=d_X(x_2,x_3) \end{align*} \begin{align*} d=d_Y(y_2,y_3). \end{align*} The triangle inequalities in $(X,d_X)$ and $(Y,d_Y)$ give \begin{align*} d_X(x_1,x_3)\le a+c \end{align*} and \begin{align*} d_Y(y_1,y_3)\le b+d. \end{align*} Since the function $t\mapsto t^p$ is increasing on $[0,\infty)$, the finite-$p$ formula gives \begin{align*} d_p((x_1,y_1),(x_3,y_3))\le \left((a+c)^p+(b+d)^p\right)^{1/p}. \end{align*} By the two-coordinate Minkowski estimate proved above, \begin{align*} \left((a+c)^p+(b+d)^p\right)^{1/p}\le \left(a^p+b^p\right)^{1/p}+\left(c^p+d^p\right)^{1/p}. \end{align*} Using the definitions of $a,b,c,d$, this becomes \begin{align*} d_p((x_1,y_1),(x_3,y_3))\le d_p((x_1,y_1),(x_2,y_2))+d_p((x_2,y_2),(x_3,y_3)). \end{align*} Thus $d_p$ satisfies the triangle inequality for every $1\le p<\infty$. [/step]

[step:Use the maximum inequality to prove the $p=\infty$ triangle inequality] Let $(x_1,y_1),(x_2,y_2),(x_3,y_3)\in X\times Y$. Define \begin{align*} a=d_X(x_1,x_2) \end{align*} \begin{align*} b=d_Y(y_1,y_2) \end{align*} \begin{align*} c=d_X(x_2,x_3) \end{align*} \begin{align*} d=d_Y(y_2,y_3). \end{align*} The coordinate triangle inequalities give \begin{align*} d_X(x_1,x_3)\le a+c \end{align*} and \begin{align*} d_Y(y_1,y_3)\le b+d. \end{align*} Therefore \begin{align*} d_\infty((x_1,y_1),(x_3,y_3))\le \max\{a+c,b+d\}. \end{align*} Since $a\le \max\{a,b\}$, $b\le \max\{a,b\}$, $c\le \max\{c,d\}$, and $d\le \max\{c,d\}$, we have \begin{align*} a+c\le \max\{a,b\}+\max\{c,d\} \end{align*} and \begin{align*} b+d\le \max\{a,b\}+\max\{c,d\}. \end{align*} Taking the maximum of the two left-hand sides gives \begin{align*} \max\{a+c,b+d\}\le \max\{a,b\}+\max\{c,d\}. \end{align*} Substituting the definitions of $a,b,c,d$ yields \begin{align*} d_\infty((x_1,y_1),(x_3,y_3))\le d_\infty((x_1,y_1),(x_2,y_2))+d_\infty((x_2,y_2),(x_3,y_3)). \end{align*} Thus $d_\infty$ satisfies the triangle inequality. [/step]

[step:Conclude that every product formula satisfies the metric axioms] For $1\le p<\infty$, the finite-$p$ formula is nonnegative, symmetric, separates points, and satisfies the triangle inequality by the preceding steps. Hence $d_p$ is a metric on $X\times Y$. For $p=\infty$, the supremum formula is nonnegative, symmetric, separates points, and satisfies the triangle inequality by the preceding steps. Hence $d_\infty$ is a metric on $X\times Y$. This proves the theorem for every $1\le p\le\infty$. [/step]