[proofplan] We prove that the usual contraction inequality is equivalent to a uniform upper bound on all distance ratios. In the forward direction, the contraction constant bounds each ratio because $d(x,y)>0$ whenever $x\ne y$, and the least-upper-bound property then bounds the supremum. In the reverse direction, a supremum bound gives the same inequality for every distinct pair, while the equal-point case follows from the metric identity $d(f(x),f(x))=0$. [/proofplan] [step:Convert a contraction constant into an upper bound for the distance-ratio supremum] Assume first that $f$ is a contraction. By definition, there exists a constant $k\in[0,1)$ such that \begin{align*} d(f(x),f(y))\le k\,d(x,y) \end{align*} for all $x,y\in X$. Define the set of distance ratios $R\subset[0,\infty)$ by \begin{align*} R:=\left\{\frac{d(f(x),f(y))}{d(x,y)}:x,y\in X,\ x\ne y\right\}. \end{align*} This set is nonempty because $X$ contains at least two distinct points. If $x,y\in X$ and $x\ne y$, then $d(x,y)>0$ by the definiteness property of a metric, so division by $d(x,y)$ gives \begin{align*} \frac{d(f(x),f(y))}{d(x,y)}\le k. \end{align*} Thus $k$ is an upper bound for $R$. Since $\sup R$ is the least upper bound of $R$ in $[0,\infty]$, we have \begin{align*} \sup R\le k. \end{align*} Taking $c:=k$ gives the required constant $c\in[0,1)$. [/step] [step:Recover the contraction inequality from the supremum bound] Conversely, assume that there exists $c\in[0,1)$ such that \begin{align*} \sup\left\{\frac{d(f(u),f(v))}{d(u,v)}:u,v\in X,\ u\ne v\right\}\le c. \end{align*} Let $x,y\in X$ be arbitrary. If $x\ne y$, then $d(x,y)>0$, and the ratio \begin{align*} \frac{d(f(x),f(y))}{d(x,y)} \end{align*} belongs to the set whose supremum is bounded above by $c$. Hence \begin{align*} \frac{d(f(x),f(y))}{d(x,y)}\le c. \end{align*} Multiplying by the positive number $d(x,y)$ gives \begin{align*} d(f(x),f(y))\le c\,d(x,y). \end{align*} If $x=y$, then $f(x)=f(y)$, so the definiteness property of the metric gives \begin{align*} d(f(x),f(y))=0=c\,d(x,y). \end{align*} Therefore \begin{align*} d(f(x),f(y))\le c\,d(x,y) \end{align*} for all $x,y\in X$. Since $c\in[0,1)$, this is exactly the definition of $f$ being a contraction. [guided] We want to prove that a bound on the supremum of all ratios gives the pointwise contraction inequality. The supremum appearing in the hypothesis is taken over the set \begin{align*} R:=\left\{\frac{d(f(u),f(v))}{d(u,v)}:u,v\in X,\ u\ne v\right\}. \end{align*} The hypothesis says $\sup R\le c$ for some $c\in[0,1)$. By the defining order property of the supremum, every element of $R$ is bounded above by $\sup R$, and therefore every element of $R$ is bounded above by $c$. Now fix arbitrary points $x,y\in X$. There are two cases because the ratio is only defined for distinct points. First suppose $x\ne y$. Since $d$ is a metric, definiteness gives $d(x,y)>0$. Therefore the ratio \begin{align*} \frac{d(f(x),f(y))}{d(x,y)} \end{align*} is a well-defined element of $R$. The supremum bound gives \begin{align*} \frac{d(f(x),f(y))}{d(x,y)}\le c. \end{align*} Because $d(x,y)>0$, multiplying both sides by $d(x,y)$ preserves the inequality, and we obtain \begin{align*} d(f(x),f(y))\le c\,d(x,y). \end{align*} It remains to handle the case $x=y$, which is not included in the ratio set. If $x=y$, then $f(x)=f(y)$ because $f:X\to X$ is a map. Applying definiteness of the metric $d$ at the point $f(x)\in X$ gives \begin{align*} d(f(x),f(y))=d(f(x),f(x))=0. \end{align*} Also $d(x,y)=d(x,x)=0$, so \begin{align*} d(f(x),f(y))=0=c\,d(x,y). \end{align*} Thus the inequality \begin{align*} d(f(x),f(y))\le c\,d(x,y) \end{align*} holds for every pair $x,y\in X$. Since the constant satisfies $c\in[0,1)$, this is precisely the contraction condition. [/guided] [/step]