= Solution
For $F\dashv U$ with <unit of an adjunction> $\eta$ and <counit of an adjunction> $\varepsilon$, put
$$
\boxed{T=UF,\qquad\mu_A=U(\varepsilon_{FA}).}
$$
The <triangle identities for an adjunction> give
$$
\mu_A\eta_{TA}=1_{TA},\qquad
\mu_AT(\eta_A)=1_{TA}.
$$
For associativity, <naturality> of $\varepsilon$ at $\varepsilon_{FA}:FUFA\to FA$ says
$$
\varepsilon_{FA}FU(\varepsilon_{FA})
=\varepsilon_{FA}\varepsilon_{FUFA}.
$$
Apply $U$ to obtain $\mu_AT(\mu_A)=\mu_A\mu_{TA}$. Thus this is the <monad induced by an adjunction>.
Conversely, start with a <monad> $(T,\eta,\mu)$. The <free algebra functor> sends $A$ to $(TA,\mu_A)$ and $f$ to $Tf$. The <monad> laws ensure that these are <algebras for a monad> and <morphisms of algebras for a monad>. For an algebra $(B,b)$, there are inverse <bijections>
$$
\mathcal C^T((TA,\mu_A),(B,b))\cong\mathcal C(A,B),
\qquad g\longmapsto g\eta_A,\quad f\longmapsto bT(f).
$$
The inverse really is an algebra map because
$$
bT(f)\mu_A=b\mu_B T^2(f)=bT(b)T^2(f)=bT(bT(f)).
$$
For a free-algebra map $g$, its algebra equation and the <monad> unit law give
$$
bT(g\eta_A)=bT(g)T(\eta_A)=g\mu_AT(\eta_A)=g.
$$
For a base map $f$, naturality and the algebra unit law give $bT(f)\eta_A=b\eta_Bf=f$. The formulas are natural, establishing the <free-forgetful Eilenberg-Moore adjunction> $F^T\dashv U^T$. Its unit is $\eta_A$, and its counit at $(B,b)$ is $b$. Hence its induced multiplication is $\mu$, and \b[every monad arises from an adjunction].
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