My understanding is that variational inequalities involve a scalar product and that equations are solved using duality relations. How come the following is a variational inequality? It rather looks like a partial differential inequality.
$L$ above is the infinitesimal generator of the underlying dynamic which is a partial differential operator.
The following is from a page in one of Bensoussan Lions books
I hope not to be too simplistic in the answer, but I share Dirk's point of view in the comments.
Variational problems arose firstly in the branch of Calculus of Variations, where to define a notion of derivative for general functionals defined on a normed vector space (or some generalizations of it) and to find local minima, one had to take "small perturbations" of the solution, trying to show the value of the functional increases. The main scheme of the variational method is indeed as follows: assuming that the optimal curve $u(x)$ exists among smooth (twice-differentiable curves), we compare the optimal curve with close-by trajectories $u(x)+\delta u(x)$, where variation $\delta u(x)$ is small in some sense. Using the smallness of $\delta u$, we simplify the comparison, deriving necessary conditions for the optimal trajectory $u(x)$. These methods are applicable to a great variety of extremal problems called, indeed, variational problems.
With the above paragraph in mind, I guess that every time we result in an inequality involving functionals that act over spaces of functions there is the temptation to see a variational method in it, hence calling it "variational inequality". I do not know what is $V$ in your case, but condition $(vi)$ is indeed showing a functional (or even more generally, an operator) acting on some function.
Just out of curiosity let me point out why calling variational inequality or partial differential equation inequality can be seen as the same thing: the Dirichlet's principle.
Dirichlet's principle states that, if the function $u(x)$ is the solution to Poisson's equation $$ \Delta u+f=0 $$ on a domain $\Omega$ of $\mathbb{R}^{n}$ with boundary condition $u=g$ on the boundary $\partial \Omega$, then $u$ can be obtained as the minimizer of the Dirichlet energy $$ I[v(x)]=\int_{\Omega}\left(\frac{1}{2}|\nabla v|^{2}-v f\right) \mathrm{d} x $$
Hence you see how to solve a PDE or to be a minimizer of a variational problem is equivalent, for some specific functional $I$. As a result, when inequalities are involves, calling them variational or PDE inequalities is not so much different.