Let $x^\star =\arg \min_{x \in C} f(x)$. Is it true that $\langle x-x^\star, \nabla f(x^\star) \rangle \ge 0$? Let $f$ be a convex function and $C$ some convex set and let 
\begin{align}
x^\star= \arg \min_{x \in C} f(x)
\end{align}  Is it true that  $\langle x-x^\star, \nabla f(x^\star) \rangle \ge 0$ for all $x\in C$. 
I was trying to show this by using the fact that if $f$ is convex, then
\begin{align}
f(u)\ge f(w) + \langle u-w, \nabla f(w) \rangle 
\end{align} 
For example, since we have that $f(x^\star)-f(x) \le 0$, then we have
\begin{align}
0 \ge \langle x^\star-x, \nabla f(x) \rangle  \Longrightarrow   \langle x-x^\star, \nabla f(x) \rangle .
\end{align} 
However, this is not the right inequality. 
 A: What you're playing with is called a Variational Inequality.
The way it works is, assume $x^*$ is a global minimizer of $f(x)$ on $C$, a convex set. Now take the derivative of 
$$
\phi(\varepsilon) = f( (1-\varepsilon)x^* + \varepsilon x')
$$
wrt to $\varepsilon$, yielding
$$
\phi'(\varepsilon) = \nabla f( (1-\varepsilon)x^* + \varepsilon x')'(x'-x^*).
$$
A necessary condition for $x^*$ to be a global minimizer is that $\phi'(0) \ge 0$, or
$$
\phi'(0) = \nabla f( x^*)'(x'-x^*) \ge 0, \quad \forall x' \in C,
$$
so that taking a directional derivative in any feasible direction $x'$ raises the value of the objective function, since $x^*$ is assumed to be the global minimizer. See how this subsumes the standard FONC's, which only apply on the interior of a set? In fact, if $x^*$ is on the interior of $C$, $\nabla f(x^*)=0$, and otherwise, $\nabla f(x^*)$ is non-zero, but characterizes the supporting hyperplane to $C$; the inequality accounts for the missing Lagrange or Kuhn-Tucker terms, $\lambda \nabla h(x^*)$ where $h$ is the constraint.
Convexity hasn't been used yet. What convexity of $f$ is buying is that the solution set of the variational inequality is a convex set. So if $x^*$ and $x'$ are global minimizers and $f$ is convex,
$$
f(\lambda x^* + (1-\lambda)x') \le \lambda f(x^*) + (1-\lambda) f(x') = f^*
$$
and the convex combination $\lambda x^* + (1-\lambda)x'$ must also be a global minimizer. If $f$ is strictly convex, that inequality would yield a contradiction, so that the minimizer is unique.
In general the VI framework generalizes standard FONC/SOSC reasoning to account for solutions on the boundary.
A: Suppose there exists $x\in C$ such that $\langle x - x^\ast, \nabla f(x^\ast)\rangle < 0$.
Since $x^\ast + \alpha (x-x^\ast)\in C$ for each $\alpha \in [0, 1)$.
Let $g(\alpha) \triangleq f(x^\ast + \alpha (x-x^\ast)) - f(x^\ast)$ for $\alpha \in [0, 1)$.
Since $f$ is differentiable at $x^\ast$, we know that $g$ is differentiable at $0$ and
$g'(0) = \langle x - x^\ast, \nabla f(x^\ast)\rangle$. 
Since $g(0) = 0$ and $g'(0) < 0$, there exists some $\alpha_0$ in $(0, 1)$ such that $g(\alpha_0) < 0$, i.e., 
$f(x^\ast + \alpha_0 (x-x^\ast)) < f(x^\ast)$ which contradicts the optimality of $x^\ast$.
We are done.
