I want to show that for a convex, 1-smooth function $f$ on a non-empty convex set $X$ it holds that: $x^*$ is a global minimum of $f$ on $X$ if and only if for all $x\in X$: $\nabla f(x^*)^T(x-x^*)\ge0$.
I got $\impliedby$: As $f$ is convex over a convex set $X$, for all $x_1, x_2\in X$ we know that $$f(x_2) \ge f(x_1)+(x_2-x_1)^T\nabla f(x_1)$$ so obviously if $(x_2-x_1)^T\nabla f(x_1) \ge 0$ then $x_1 \le x_2$.
I can't get $\implies$ though, I don't think it follows from the same equation. Can anyone offer a suggestion?
By contradiction: Let $x^\ast \in X$ be a global minimum of $f$ and assume there exists an $y \in X$ s.t. $\nabla f(x^\ast)^T (y - x^\ast) < 0.$ Let $z(t) = (1-t)x^\ast + ty$ for $t \in [0, 1].$ Then $$\frac{d}{dt} f(z(t)) = \nabla_zf(z(t))^T (y - x^\ast). $$
At $t=0$ we have $z(t) = x^\ast$, so $$ \frac{d}{dt}\big|_{t=0} f(z(t)) = \nabla f(x^\ast)^T (y - x^\ast) < 0, $$ so there exists a point $a$ in the neighborhood of $x^\ast$ with $f(a) < f(x^\ast).$