Let $\mathcal{D} \underset{G}{\overset{F}{\leftrightarrows}}\mathcal{C}$ with $F\dashv G$ and $G$ fully faithful. I want to show that this implies that the counit is an isomorphism. I was trying to do this in a similar way as in this answer to a similar question. I ran into some issues. This is what I tried:
$$\mathcal{D}(A,B)\cong \mathcal{C}(GA,GB)\cong\mathcal{D}(FGA,B)$$
(First natural iso from fully faithfulness of $G$ and second natural iso from adjunction). Now by a corollary of the Yoneda Lemma there is a natural isomorphism $i:A\overset{\cong}{\rightarrow}FGA$. I would like to show that this is precisely the component $\epsilon_A$ of the counit of the adjunction. A way to do it would be to find $\eta$ such that $\eta,i$ respect the triangle inequalities. It seems that a good candidate for $\eta_X$ would be the image of the identity under the natural isomorphism $\mathcal{D}(FX,FX)\rightarrow \mathcal{C}(X,GFX)$.
But I failed to check that the triangle identities hold, and I'm afraid I chose the unit or the counit wrongly.