TITLE: Prove this basic inequality $f(y)\geq f(x)+\nabla f(x)^T(y-x)$for differentiable convex functions
QUESTION [5 upvotes]: Prove this basic inequality $f(y)\geq f(x)+\nabla f(x)^T(y-x)$for differentiable convex functions.
I tried using Taylor Expansion to get
$$f(y)=f(x)+\nabla f(x)^T(y-x)+\frac{1}{2}(y-x)^TH_f(x)(y-x)+\dots$$
Can I simply claim that the inequality follows? How?

REPLY [5 votes]: Consider the function of $t\in[0,1]$:
$$
g(t)=f(x(1-t)+yt)\tag{1}
$$
Then apply the mean value theorem: there exists a $t\in(0,1)$ so that
$$
g(1)-g(0)=g'(t)\ge g'(0)\tag{2}
$$
The last inequality is because $g$ is convex.
Write $(2)$ in terms of $f$:
$$
f(y)-f(x)\ge\nabla f(x)\cdot(y-x)\tag{3}
$$