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A straightforward estimates show that any $n$-point $\delta$-hyperbolic space lies on distance at most $n{\cdot}\delta$ from a $0$-hyperbolic space. The following statement improves this bound to $\log n\cdot \delta$. I learned it from Rostislav Matveyev, who attributed it to Étienne Ghys.

Let $d$ be a $\delta$-hyperbolic metric on an $n$-points set $F$. Then there is a $0$-hyperbolic metric $d'$ on $F$ such that $$d \leqslant d' \leqslant d+\mathrm{const}\cdot\log n\cdot \delta.$$

To prove it, you choose a base point and define new Gromov's product $(\ |\ )'$ using the old one $(\ |\ )$ $$(x|y)'\mathrel{:=}\max \{\,\min \{\,(x_0|x_1),\dots,(x_{m-1}|x_m)\,\}\,\},$$ where the maximum is taken for all chains of points $x=x_0,\dots, x_m=y$. Plus you need to apply the following lemma:

For any sequence of points $x_0,\dots,x_m$ in a $\delta$-hyperbolic space, we have $$(x_0|x_m)\geqslant \min \{\,(x_0|x_1),\dots,(x_{m-1}|x_m)\,\}-\mathrm{const}\cdot\log m\cdot \delta.$$

The proof is recursive application of the inequality for Gromov's product for triples $$(x_i|x_k)\geqslant \min\{\,(x_{i}|x_j),(x_{j}|x_k)\,\}- \delta.$$ You start with $(x_0|x_m)$ and choose roughly the middle index each time. This way you get $\lceil\log_2 n\rceil$ iterations; each iteration gives you extra $\delta$.

Postscript. Actually, it is Theorem 12 in "Espaces métriques hyperboliques" by Étienne Ghys and Pierre de la Harpe.

A straightforward estimates show that any $n$-point $\delta$-hyperbolic space lies on distance at most $n{\cdot}\delta$ from a $0$-hyperbolic space. The following statement improves this bound to $\log n\cdot \delta$. I learned it from Rostislav Matveyev, who attributed it to Étienne Ghys.

Let $d$ be a $\delta$-hyperbolic metric on an $n$-points set $F$. Then there is a $0$-hyperbolic metric $d'$ on $F$ such that $$d \leqslant d' \leqslant d+\mathrm{const}\cdot\log n\cdot \delta.$$

To prove it, you choose a base point and define new Gromov's product $(\ |\ )'$ using the old one $(\ |\ )$ $$(x|y)'\mathrel{:=}\max \{\,\min \{\,(x_0|x_1),\dots,(x_{m-1}|x_m)\,\}\,\},$$ where the maximum is taken for all chains of points $x=x_0,\dots, x_m=y$. Plus you need to apply the following lemma:

For any sequence of points $x_0,\dots,x_m$ in a $\delta$-hyperbolic space, we have $$(x_0|x_m)\geqslant \min \{\,(x_0|x_1),\dots,(x_{m-1}|x_m)\,\}-\mathrm{const}\cdot\log m\cdot \delta.$$

The proof is recursive application of the inequality for Gromov's product for triples $$(x_i|x_k)\geqslant \min\{\,(x_{i}|x_j),(x_{j}|x_k)\,\}- \delta.$$ You start with $(x_0|x_m)$ and choose roughly the middle index each time. This way you get $\lceil\log_2 n\rceil$ iterations; each iteration gives you extra $\delta$.

Postscript. Actually, it is 6.1.B in Gromov's "Hyperbolic groups".

A straightforward estimates show that any $n$-point $\delta$-hyperbolic space lies on distance at most $n{\cdot}\delta$ from a $0$-hyperbolic space. The following statement improves this bound to $\log n\cdot \delta$. I learned it from Rostislav Matveyev, who attributed it to Étienne Ghys.

Let $d$ be a $\delta$-hyperbolic metric on an $n$-points set $F$. Then there is a $0$-hyperbolic metric $d'$ on $F$ such that $$d \leqslant d' \leqslant d+\mathrm{const}\cdot\log n\cdot \delta.$$

To prove it, you choose a base point and define new Gromov's product $(\ |\ )'$ using the old one $(\ |\ )$ $$(x|y)'\mathrel{:=}\max \{\,\min \{\,(x_0|x_1),\dots,(x_{m-1}|x_m)\,\}\,\},$$ where the maximum is taken for all chains of points $x=x_0,\dots, x_m=y$. Plus you need to apply the following lemma:

For any sequence of points $x_0,\dots,x_m$ in a $\delta$-hyperbolic space, we have $$(x_0|x_m)\geqslant \min \{\,(x_0|x_1),\dots,(x_{m-1}|x_m)\,\}-\mathrm{const}\cdot\log m\cdot \delta.$$

The proof is recursive application of the inequality for Gromov's product for triples $$(x_i|x_k)\geqslant \min\{\,(x_{i}|x_j),(x_{j}|x_k)\,\}- \delta.$$ You start with $(x_0|x_m)$ and choose roughly the middle index each time. This way you get $\lceil\log_2 n\rceil$ iterations; each iteration gives you extra $\delta$.

A straightforward estimates show that any $n$-point $\delta$-hyperbolic space lies on distance at most $n{\cdot}\delta$ from a $0$-hyperbolic space. The following statement improves this bound to $\log n\cdot \delta$. I learned it from Rostislav Matveyev, who attributed it to Étienne Ghys.

Let $d$ be a $\delta$-hyperbolic metric on an $n$-points set $F$. Then there is a $0$-hyperbolic metric $d'$ on $F$ such that $$d \leqslant d' \leqslant d+\mathrm{const}\cdot\log n\cdot \delta.$$

To prove it, you choose a base point and define new Gromov's product $(\ |\ )'$ using the old one $(\ |\ )$ $$(x|y)'\mathrel{:=}\max \{\,\min \{\,(x_0|x_1),\dots,(x_{m-1}|x_m)\,\}\,\},$$ where the maximum is taken for all chains of points $x=x_0,\dots, x_m=y$. Plus you need to apply the following lemma:

For any sequence of points $x_0,\dots,x_m$ in a $\delta$-hyperbolic space, we have $$(x_0|x_m)\geqslant \min \{\,(x_0|x_1),\dots,(x_{m-1}|x_m)\,\}-\mathrm{const}\cdot\log m\cdot \delta.$$

The proof is recursive application of the inequality for Gromov's product for triples $$(x_i|x_k)\geqslant \min\{\,(x_{i}|x_j),(x_{j}|x_k)\,\}- \delta.$$ You start with $(x_0|x_m)$ and choose roughly the middle index each time. This way you get $\lceil\log_2 n\rceil$ iterations; each iteration gives you extra $\delta$.

Postscript. Actually, it is Theorem 12 in "Espaces métriques hyperboliques" by Étienne Ghys and Pierre de la Harpe.

A straightforward estimates show that any $n$-points $\delta$-hyperbolic lies on distance at most $n{\cdot}\delta$ from a $0$-hyperbolic space. The following statement improves this bound to $\log n\cdot \delta$. I learned it from Rostislav Matveyev, who attributed it to Étienne Ghys.

Let $d$ be a $\delta$-hyperbolic metric on an $n$-points set $F$. Then there is a $0$-hyperbolic metric $d'$ on $F$ such that $$d \leqslant d' \leqslant d+\mathrm{const}\cdot\log n\cdot \delta.$$

To prove it, you choose a base point and define new Gromov's product $(\ |\ )'$ using the old one $(\ |\ )$ $$(x|y)'\mathrel{:=}\max \{\,\min \{\,(x_0|x_1),\dots,(x_{m-1}|x_m)\,\}\,\},$$ where the maximum is taken for all chains of points $x=x_0,\dots, x_m=y$. Plus you need to apply the following lemma:

For any sequence of points $x_0,\dots,x_m$ in a $\delta$-hyperbolic space, we have $$(x_0|x_m)\geqslant \min \{\,(x_0|x_1),\dots,(x_{m-1}|x_m)\,\}-\mathrm{const}\cdot\log m\cdot \delta.$$

The proof is recursive application of the inequality for Gromov's product for triples $$(x_i|x_k)\geqslant \min\{\,(x_{i}|x_j),(x_{j}|x_k)\,\}- \delta.$$ You start with $(x_0|x_m)$ and choose roughly the middle index each time. This way you get $\lceil\log_2 n\rceil$ iterations; each iteration gives you extra $\delta$.

A straightforward estimates show that any $n$-point $\delta$-hyperbolic space lies on distance at most $n{\cdot}\delta$ from a $0$-hyperbolic space. The following statement improves this bound to $\log n\cdot \delta$. I learned it from Rostislav Matveyev, who attributed it to Étienne Ghys.

Let $d$ be a $\delta$-hyperbolic metric on an $n$-points set $F$. Then there is a $0$-hyperbolic metric $d'$ on $F$ such that $$d \leqslant d' \leqslant d+\mathrm{const}\cdot\log n\cdot \delta.$$

To prove it, you choose a base point and define new Gromov's product $(\ |\ )'$ using the old one $(\ |\ )$ $$(x|y)'\mathrel{:=}\max \{\,\min \{\,(x_0|x_1),\dots,(x_{m-1}|x_m)\,\}\,\},$$ where the maximum is taken for all chains of points $x=x_0,\dots, x_m=y$. Plus you need to apply the following lemma:

For any sequence of points $x_0,\dots,x_m$ in a $\delta$-hyperbolic space, we have $$(x_0|x_m)\geqslant \min \{\,(x_0|x_1),\dots,(x_{m-1}|x_m)\,\}-\mathrm{const}\cdot\log m\cdot \delta.$$

The proof is recursive application of the inequality for Gromov's product for triples $$(x_i|x_k)\geqslant \min\{\,(x_{i}|x_j),(x_{j}|x_k)\,\}- \delta.$$ You start with $(x_0|x_m)$ and choose roughly the middle index each time. This way you get $\lceil\log_2 n\rceil$ iterations; each iteration gives you extra $\delta$.