Supporting inequality for a differentiable convex function ($f(y)\ge f(x)+\nabla f(x{)}^{\mathsf{T}}(y-x)$)

For a differentiable real-valued function on an open convex set, convexity is equivalent to the displayed inequality for all $x,y$ in its domain. To prove the forward direction, convexity gives $[f(x+t(y-x))-f(x)]/t\le f(y)-f(x)$ for $0<t\le 1$. Take the limit as $t\downarrow 0$ using first-order Taylor expansion.

Conversely, put $z=(1-t)x+ty$ and apply the supporting inequality at $z$ to both $x$ and $y$. Multiply the two inequalities by $1-t$ and $t$ and add them. The gradient terms cancel because $(1-t)(x-z)+t(y-z)=0$, leaving $f(z)\le (1-t)f(x)+tf(y)$. This is the defining convexity inequality. The endpoint values $t=0,1$ hold as equalities.