The following is an excerpt of a proof about monotone operators, I hope not too much context is required because I only need to understand an inequality:
Preconditions: $V$ is a normed space, $A:V \to V^*$ is a coercive, monotone operator.
Excerpt from the proof:
$... \geq \int_\frac{1}{2}^1 \big\langle A(tv)-A(0), v \big\rangle \,\mathrm{d}t \geq \frac{1}{2} \big\langle A(\frac{1}{2}v)-A(0),v \big\rangle$
How does one obtain the last inequality?
This inequality follows by integration of $$ \langle A(t\,v) - A(0), v \rangle \ge A(\frac12 \, v) - A(0), v \rangle$$ for all $t \ge \frac12$. Note that this inequality is equivalent to $$ \langle A(t \, v) - A(\frac12 \, v) , v \rangle = \frac1{t - \frac12} \, \langle A(t\,v) - A(\frac12\, v), t \, v - \frac12 \, v\rangle\ge 0$$ for all $ t \ge \frac12$ which follows from monotonicity.