Prove this basic inequality $f(y)\geq f(x)+\nabla f(x)^T(y-x)$for differentiable convex functions

Question

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?

Answer 1

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 $f$: $$ f(y)-f(x)\ge\nabla f(x)\cdot(y-x)\tag{3} $$