Definition. Let $X,Y$ metric spaces. A map $f:X\to Y$ is called $(L,C)$-coarse Lipschitz if $$d_Y(f(x),f(x'))\leq Ld(x,x')+C$$ for all $x,x'\in X$.
A map $f:X\to Y$ is called $(L,C)$-quasiisometric embedding if $$\frac{1}{L}d_X(x,x')-C\leq d_Y(f(x),f(x'))\leq Ld_X(x,x')+C$$ for all $x,x'\in X$
Definition. Maps of metric spaces $f:X\to Y,\ \overline{f}:Y\to X$ are said to be $C$-coarse inverse to each other if $$d_X(\overline{f}\circ f,Id_X)\leq C,\quad d_Y(f\circ\overline{f},Id_Y)\leq C$$
Definition. A map $f:X\to Y$ between metric spaces is called a quasiisometry if it is coarse Lipschitz and admits a coarse Lipschitz coarse inverse map. More precisely, $f$ is an $(L,C)$-quasiisometry if f is $(L,C)$-coarse Lipschitz and there exists an $(L,C)$-coarse Lipschitz map $\overline{f}:Y\to X$ such that the maps $f,\overline{f}$ are $C$-coarse inverse to each other. Two metric spaces $X,Y$ are quasiisometrics if there exists a quasiisometry $X\to Y$.
How prove that every quasiisometry $f:X\to Y$ is a quasiisometric embedding? I have this: Let $f:X\to Y$ quasiisometry then $d_Y(f(x),f(x'))\leq Ld_X(x,x')+C$ And, $$d_X(\overline{f}(y),\overline{f}(y'))\leq Ld_Y(y,y')+C$$ then $$\frac{1}{L}d_X(\overline{f}(y),\overline{f}(y'))-C\leq d_Y(y,y')$$
If we could put $y=f(x)$ and $y'=f(x')$ (but f is not necessary surjective...) then $$\frac{1}{L}d_X(\overline{f}f(x),\overline{f}f(x'))-C\leq d_Y(f(x),f(x'))$$
The trick is to start with $\bar{f}f(x)$ and $\bar{f}f(x')$, rather than $y$ and $y'$. Using first the fact that $\bar{f}$ is coarsely Lipschitz, then that $f$ is, leads to: $$ d_X(\bar{f}f(x),\bar{f}f(x'))\leq Ld_Y(f(x),f(x'))+C\leq L^2d_X(x,x')+LC+C $$ Since $f$ and $\bar{f}$ are coarse inverses, applying the triangle inequality to the left-hand side gives the result.