I am trying to prove a lemma in Burago's "A Course in Metric Spaces" (Exercise 8.4.4, p.286). Here is a link to a different person's question about the very next exercise in that book, which also contains the definitions of a geodesic metric space and the Gromov's product $(a,b)_c$.
A geodesic metric space $(X,d)$ is $\delta$-hyperbolic if for any $a,b,c\in X$ the shortest path $[ac]$ is contained in the $\delta$-neighbourhood of $[ab]\bigcup[bc]$.
The exercise I am struggling with goes as follows:
Let $[ab],[ac]$ be two shortest paths. Let $b'\in[ab]$ and $c'\in[ac]$ be such that $d(a,b')=d(a,c')\le(b,c)_a$. Then $d(b',c')\le 3\delta$. Moreover, there is a point $p\in[bc]$ such that $[bp]$ lies in $\delta$-neighbourhood of $[ba]$ and $[cp]$ lies in $\delta$-neighbourhood of $[ca]$.
I can show that $d(b',c')\le 4\delta$, and as for the second claim I'm out of ideas.
The book suggests to use a similar reasoning to the proof that $(b,c)_a\le d(a,[bc])\le (b,c)_a+2\delta$, but I feel that copying it here will make the question too lengthy.
UPD: while proceeding with studying the subject I have stumbled upon another exercise that I cannot solve, from a different book. Since there is no answers for my original question so far, I'll add the second one here. Instead of quoting the exercise, I'll better discuss the result, it most likely originates from, namely 6.6.A. from Gromov's Hyperbolic Groups (Essays in group theory, 75–263).
That one basically says that if we were able to find points $\overline{y}_1,\overline{y}_2,\overline{y}_3$ on the corresponding sides of the geodesic triangle $[e_1,e_2,e_3]$, such that all the distances between them is less than $\delta$, then for the points $y_i$ on the corresponding sides, such that $d(x_1,y_2)=d(x_1,y_3), d(x_2,y_3)=d(x_2,y_3), d(x_3,y_2)=d(x_3,y_1)$ , then all the distances between $y_i$ is less than $2\delta$.
From the (inverse) triangle inequality we get that $|d(x_1,\overline{y}_2)-d(x_1,\overline{y}_3)|<\delta$ (and so on), from which "by easy computation" it follows that $|d(x_1,y_2)-d(x_1,\overline{y}_2)|<\frac{1}{2}\delta$. Could you please explain these easy computations? I thought I understood what that means, but I ended up with just on of the three distances being less then $\frac{1}{2}\delta$, not all of them. In fact, I think, that just from computing the segments of the triangle it is impossible to deduce the statement. Am I wrong?
I can however prove the statement with $3\delta$, instead of $2\delta$, so it's not vital.
Thank you.
