= Solution
Using the <dual pairing> between $X^*$ and $X$, a <monotone operator> satisfies
$$
\boxed{\langle A(u)-A(v),u-v\rangle\geq0\quad(u,v\in M).}
$$
A <nonlinear coercive operator> with respect to $u_0$ satisfies
$$
\boxed{\frac{\langle A(u),u-u_0\rangle}{\|u-u_0\|}\longrightarrow+\infty\quad\text{as }u\in M,\ \|u-u_0\|\to\infty.}
$$
The <directional derivative> is $\partial_vF(u)=\lim_{t\to0}[F(u+tv)-F(u)]/t$, when this finite limit exists through admissible points. At a boundary point of $M$ only an inward direction may be admissible; then one uses the corresponding one-sided limit.
If $A:X\to X^*$ is monotone and $s<t$, apply its defining inequality at $u+tv$ and $u+sv$. It becomes $(t-s)[\varphi(t)-\varphi(s)]\geq0$, so $\varphi$ is nondecreasing. Conversely choose $u=y$, $v=x-y$. Nondecrease from zero to one gives $\langle A(x)-A(y),x-y\rangle\geq0$. This proves the <line criterion for a monotone operator>.
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