[proofplan] We verify the metric axioms directly from the coordinate formula. Nonnegativity and symmetry follow from elementary properties of absolute value, while identity of indiscernibles follows because a finite sum of nonnegative [real numbers](/page/Real%20Numbers) vanishes only when every summand vanishes. The only substantive point is the triangle inequality; we prove the finite-dimensional Cauchy-Schwarz estimate inside the argument and then apply it to the coordinate differences $x-z$ and $z-y$. [/proofplan]

[step:Verify that $d_2$ is nonnegative] Fix $n\in\mathbb{N}$. Let $x=(x_1,\dots,x_n)\in\mathbb{R}^n$ and $y=(y_1,\dots,y_n)\in\mathbb{R}^n$. For each index $k\in\{1,\dots,n\}$, the real number $|x_k-y_k|^2$ is nonnegative. Hence the finite sum \begin{align*} \sum_{k=1}^n |x_k-y_k|^2 \end{align*} is nonnegative, so its square root is defined and nonnegative. Therefore \begin{align*} d_2(x,y)\geq 0. \end{align*} [/step]

[step:Show that zero distance is equivalent to equality of points]Let $x=(x_1,\dots,x_n)\in\mathbb{R}^n$ and $y=(y_1,\dots,y_n)\in\mathbb{R}^n$. If $x=y$, then $x_k=y_k$ for every $k\in\{1,\dots,n\}$, and therefore \begin{align*} d_2(x,y)=\left(\sum_{k=1}^n 0^2\right)^{1/2}=0. \end{align*} Conversely, suppose $d_2(x,y)=0$. Squaring both sides gives \begin{align*} \sum_{k=1}^n |x_k-y_k|^2=0. \end{align*} Each summand $|x_k-y_k|^2$ is nonnegative. A finite sum of nonnegative real numbers is zero only if every summand is zero, so $|x_k-y_k|^2=0$ for every $k\in\{1,\dots,n\}$. Hence $|x_k-y_k|=0$, so $x_k=y_k$ for every $k\in\{1,\dots,n\}$. Thus $x=y$.[/step]

[guided]We must prove both directions because the identity of indiscernibles for a metric says exactly that $d_2(x,y)=0$ happens precisely when $x=y$. First assume $x=y$. Equality of vectors in $\mathbb{R}^n$ means equality of every coordinate, so $x_k=y_k$ for each $k\in\{1,\dots,n\}$. Thus every coordinate difference is zero, and the defining formula gives \begin{align*} d_2(x,y)=\left(\sum_{k=1}^n |x_k-y_k|^2\right)^{1/2}=\left(\sum_{k=1}^n 0^2\right)^{1/2}=0. \end{align*} Now assume $d_2(x,y)=0$. Since $d_2(x,y)$ is the nonnegative square root of the sum of squares, squaring the equality $d_2(x,y)=0$ gives \begin{align*} \sum_{k=1}^n |x_k-y_k|^2=0. \end{align*} For every index $k$, the number $|x_k-y_k|^2$ is nonnegative. If one of these summands were positive, then the whole finite sum would be positive. Therefore every summand must be zero: \begin{align*} |x_k-y_k|^2=0 \end{align*} for every $k\in\{1,\dots,n\}$. A square of a real number is zero only when the real number itself is zero, so $|x_k-y_k|=0$. Absolute value is zero only at zero, hence $x_k-y_k=0$, and therefore $x_k=y_k$ for every $k$. Equality of all coordinates gives $x=y$.[/guided]

[step:Prove symmetry from symmetry of absolute value] Let $x=(x_1,\dots,x_n)\in\mathbb{R}^n$ and $y=(y_1,\dots,y_n)\in\mathbb{R}^n$. For every $k\in\{1,\dots,n\}$, \begin{align*} |x_k-y_k|=|-(y_k-x_k)|=|y_k-x_k|. \end{align*} Squaring and summing over $k$ gives \begin{align*} \sum_{k=1}^n |x_k-y_k|^2=\sum_{k=1}^n |y_k-x_k|^2. \end{align*} Taking nonnegative square roots yields \begin{align*} d_2(x,y)=d_2(y,x). \end{align*} [/step]

[step:Establish the finite-dimensional Cauchy-Schwarz estimate][claim:Finite-dimensional Cauchy-Schwarz estimate] For all real numbers $a_1,\dots,a_n,b_1,\dots,b_n$, \begin{align*} \left|\sum_{k=1}^n a_k b_k\right|\leq \left(\sum_{k=1}^n a_k^2\right)^{1/2}\left(\sum_{k=1}^n b_k^2\right)^{1/2}. \end{align*} [/claim] [proof] Define \begin{align*} A:=\sum_{k=1}^n a_k^2 \end{align*} and \begin{align*} B:=\sum_{k=1}^n b_k^2. \end{align*} If $B=0$, then each $b_k^2=0$, so each $b_k=0$, and both sides of the desired inequality are zero. Assume $B>0$. Define the real quadratic polynomial $q:\mathbb{R}\to\mathbb{R}$ by \begin{align*} q(t):=\sum_{k=1}^n (a_k+t b_k)^2. \end{align*} For every $t\in\mathbb{R}$, $q(t)\geq 0$. Expanding the square gives \begin{align*} q(t)=A+2t\sum_{k=1}^n a_k b_k+t^2B. \end{align*} Since this quadratic polynomial has leading coefficient $B>0$ and is nonnegative for all real $t$, its discriminant is nonpositive: \begin{align*} 4\left(\sum_{k=1}^n a_k b_k\right)^2-4AB\leq 0. \end{align*} Therefore \begin{align*} \left(\sum_{k=1}^n a_k b_k\right)^2\leq AB. \end{align*} Taking nonnegative square roots gives the desired inequality. [/proof] This proves the finite-dimensional Cauchy-Schwarz estimate needed for the triangle inequality.[/step]

[guided]The triangle inequality will require us to control a cross term of the form $\sum_{k=1}^n a_k b_k$. The needed estimate is the finite-dimensional [Cauchy-Schwarz inequality](/theorems/432), and we prove it here from the nonnegativity of squares. Let $a_1,\dots,a_n,b_1,\dots,b_n$ be real numbers. Define \begin{align*} A:=\sum_{k=1}^n a_k^2 \end{align*} and \begin{align*} B:=\sum_{k=1}^n b_k^2. \end{align*} If $B=0$, then the finite sum of nonnegative terms $\sum_{k=1}^n b_k^2$ is zero, so $b_k=0$ for every $k$. Hence $\sum_{k=1}^n a_kb_k=0$, and the estimate becomes $0\leq 0$. Now assume $B>0$. Define a function \begin{align*} q:\mathbb{R}\to\mathbb{R} \end{align*} by \begin{align*} q(t):=\sum_{k=1}^n (a_k+t b_k)^2. \end{align*} For every real number $t$, each term $(a_k+t b_k)^2$ is nonnegative, so $q(t)\geq 0$. Expanding the square inside the finite sum gives \begin{align*} q(t)=\sum_{k=1}^n a_k^2+2t\sum_{k=1}^n a_kb_k+t^2\sum_{k=1}^n b_k^2. \end{align*} With the definitions of $A$ and $B$, this is \begin{align*} q(t)=A+2t\sum_{k=1}^n a_kb_k+t^2B. \end{align*} Why does this imply an inequality? A real quadratic polynomial with positive leading coefficient that is nonnegative for all real $t$ cannot have two distinct real roots. Equivalently, its discriminant must be nonpositive. Applying this to $q$ gives \begin{align*} 4\left(\sum_{k=1}^n a_kb_k\right)^2-4AB\leq 0. \end{align*} After dividing by $4$, we obtain \begin{align*} \left(\sum_{k=1}^n a_kb_k\right)^2\leq AB. \end{align*} Both sides are nonnegative, so taking square roots gives \begin{align*} \left|\sum_{k=1}^n a_kb_k\right|\leq A^{1/2}B^{1/2}. \end{align*} Substituting the definitions of $A$ and $B$ yields \begin{align*} \left|\sum_{k=1}^n a_k b_k\right|\leq \left(\sum_{k=1}^n a_k^2\right)^{1/2}\left(\sum_{k=1}^n b_k^2\right)^{1/2}. \end{align*}[/guided]

[step:Prove the triangle inequality by expanding the squared distance] Let $x=(x_1,\dots,x_n)\in\mathbb{R}^n$, $y=(y_1,\dots,y_n)\in\mathbb{R}^n$, and $z=(z_1,\dots,z_n)\in\mathbb{R}^n$. Define real numbers \begin{align*} a_k:=x_k-z_k \end{align*} and \begin{align*} b_k:=z_k-y_k \end{align*} for each $k\in\{1,\dots,n\}$. Then $x_k-y_k=a_k+b_k$ for every $k$. Hence \begin{align*} d_2(x,y)^2=\sum_{k=1}^n |a_k+b_k|^2. \end{align*} Since $a_k+b_k$ is real, $|a_k+b_k|^2=(a_k+b_k)^2$, so \begin{align*} d_2(x,y)^2=\sum_{k=1}^n a_k^2+2\sum_{k=1}^n a_kb_k+\sum_{k=1}^n b_k^2. \end{align*} By the finite-dimensional Cauchy-Schwarz estimate, \begin{align*} \sum_{k=1}^n a_kb_k\leq \left(\sum_{k=1}^n a_k^2\right)^{1/2}\left(\sum_{k=1}^n b_k^2\right)^{1/2}. \end{align*} Therefore \begin{align*} d_2(x,y)^2\leq \left(\sum_{k=1}^n a_k^2\right)+2\left(\sum_{k=1}^n a_k^2\right)^{1/2}\left(\sum_{k=1}^n b_k^2\right)^{1/2}+\left(\sum_{k=1}^n b_k^2\right). \end{align*} The right-hand side is the square \begin{align*} \left(\left(\sum_{k=1}^n a_k^2\right)^{1/2}+\left(\sum_{k=1}^n b_k^2\right)^{1/2}\right)^2. \end{align*} Using the definitions of $a_k$ and $b_k$, this becomes \begin{align*} d_2(x,y)^2\leq \left(d_2(x,z)+d_2(z,y)\right)^2. \end{align*} Both sides are squares of nonnegative real numbers, so taking nonnegative square roots gives \begin{align*} d_2(x,y)\leq d_2(x,z)+d_2(z,y). \end{align*} [/step]

[step:Conclude that all metric axioms hold] For arbitrary $x,y,z\in\mathbb{R}^n$, we have proved nonnegativity, identity of indiscernibles, symmetry, and the triangle inequality: \begin{align*} d_2(x,y)\geq 0. \end{align*} \begin{align*} d_2(x,y)=0 \iff x=y. \end{align*} \begin{align*} d_2(x,y)=d_2(y,x). \end{align*} \begin{align*} d_2(x,y)\leq d_2(x,z)+d_2(z,y). \end{align*} These are exactly the metric axioms on $\mathbb{R}^n$. Since $n\in\mathbb{N}$ was arbitrary, $d_2$ is a metric on $\mathbb{R}^n$ for every $n\in\mathbb{N}$. [/step]