Do adjoint functors really define monads? It is often claimed as "obvious" that a pair of adjoint functors: $L\colon{\cal V}\to {\cal M}$ and $R\colon{\cal M}\to {\cal V}$ defines a cotriple $(\bot, \epsilon, \delta)$ and a monad. What is wrong with the following counterexample?
Let ${\cal V}$ be the category of vector spaces over $\mathbb R$ and ${\cal M}$ be the category of modules over a Lie algebra $\mathfrak g.$ Then (I believe) the forgetful functor $R:{\cal M}\to {\cal V}$ is right adjoint to $L\colon{\cal V}\to {\cal M},$ $L(V)=V\otimes \mathfrak g.$ 
The counit $\epsilon\colon M\otimes \mathfrak g\to M$ is given by the action of $\mathfrak g$ on $M.$ In Weibel's "Intro to Homological Algebra", the construction of a monad is based on the following identity: $\epsilon\circ (LR\epsilon)=\epsilon\circ(\epsilon LR)$. That means that $(m\cdot x)\cdot y=m\cdot (x\cdot y)$ for $x,y\in \mathfrak g.$ But for Lie algebra modules we have $(m\cdot x)\cdot y=m\cdot (x\cdot y)+(m\cdot y)\cdot x$!

I would leave this as a comment, but see Mac Lane's "Categories for the Working Mathematician", second edition, p.138. He does not leave it at "it's obvious", but explains very clearly.
EDIT: Also, it took me a bit to notice this, but draw out the identity $\epsilon\circ LR\epsilon = \epsilon\circ \epsilon LR$ as a commutative diagram and observe that it's a naturality square for $\epsilon$. If your "counterexample" holds, $\epsilon$ isn't even a natural transformation, and thus not the counit of an adjunction.
