Suppose we have an adjoint triple $F \dashv G \dashv H$ with the following (co)units: $$\eta : I \to GF, \ \epsilon \colon FG \to I, \ \bar{\eta} : I \to HG, \ \bar{\epsilon} \colon GH \to I.$$ Consider the diagram
$$\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} FGH & \ra{F \bar{\epsilon}} & F \\ \da{\epsilon H} & & \da{\bar{\eta} F} \\ H & \ras{H \eta} & HGF \end{array}$$
Question: Is this diagram commutative?
If necessary, we can assume that $G$ is fully faithfull, which implies that $\bar{\eta}$ and $\epsilon$ are isomorphisms.
I belive it might be possible to construct cube-diagram from triagle identities and naturality conditions on (co)units, such that uper base of that cube is our considered diagram, but I haven't been able to do that.