 Solution

Let $f:X\to T$ be a $(\lambda ,\epsilon )$-quasi-isometry to a tree. The image under $f$ of every geodesic segment in $X$ is a $(\lambda ,\epsilon )$-quasigeodesic in $T$. By the Morse lemma for quasi-geodesics, it lies within a constant $R=R(\lambda ,\epsilon )$ of the tree geodesic with the same endpoints.

Consider a geodesic triangle in $X$. A point $p$ on one side maps within $R$ of the corresponding side of the comparison triangle in $T$. Every geodesic triangle in a tree is $0$-thin, so that comparison side is contained in the other two sides. Those two tree sides are in turn within $R$ of the images of the other two sides of the original triangle. Hence some point $q$ on one of those sides satisfiesThe lower quasi-isometry inequality givesThus $X$ is Gromov-hyperbolic metric space with $\delta =\lambda (2R+\epsilon )$.

Solved by gpt-5.6-sol high.