Obtaining orthogonality from a variational inequality in $L^2$

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I'm working on the following problem:

Let $\Omega$ be a bounded domain in $\mathbb{R}^n$ and let $H$ be a closed linear subspace of $L^2(\Omega)$. Let $\gamma : \mathbb R \to \mathbb R$ be a continuous increasing function satisfying $|\gamma(t)| \leq A + B|t|$ for some constants $A,B$. Let $F \in L^2(\Omega)$.

Show that there exists a unique $u \in H$ such that $u + \gamma \circ u -F \in H^{\perp} $.

(Hint: solve the variational inequality $(u + \gamma \circ u -F,v-u) \geq 0 \quad \forall v \in H $.)

Question: if $u \in H$ is a solution of the inequality, why does this imply $u + \gamma \circ u -F \in H^{\perp}$?

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We assume that $u \in H$ is such that $(u + \gamma \circ u -F,v-u) \geq 0$ for every $v \in H$. In particular, this means that for $w \in H$ we can take $v = u \pm w$ to obtain $(u + \gamma \circ u -F,\pm w) \geq 0$ which in turn implies that $(u + \gamma \circ u -F, w) = 0$. Since $w$ was arbitrary, this shows that $u + \gamma \circ u -F \in H^\perp$.