Let $f\in C^1(\mathbb R)$ and assume $f'$ is Lipschitz continuous with Lipschitz constant $c\ge 0$. How can we show that $$f(y)-f(x)-f'(x)(y-x)\ge-\frac c2(y-x)^2\tag1$$ for all $x,y\in\mathbb R$?
I don't understand where the factor $\frac12$ on the right-hand side comes from. And I think we need to assume $x\le y$, don't we?
If $x\le y$, then (by the mean value theorem) there is a $z\in[x,y]$ with $$f(y)-f(x)=f'(z)(y-x)=\left(f'(z)-f'(z)+f'(x)\right)(y-x)\tag2.$$ By Lipschitz continuity, $$\left|f'(z)-f'(x)\right|\le c(z-x)\le c(y-x)\tag3$$ and hence $$f'(z)-f'(x)\ge-c(y-x).\tag4$$ Thus, $$f(y)-f(x)-f'(x)(y-x)=\left(f'(z)-f'(x)\right)(y-x)\ge-c(y-x)^2.\tag5$$
How can we obtain the stronger conclusion $(1)$?
We have that $$f(y)=f(x)+\int_x^y f'(t)dt=f(x)+f'(x)(y-x)+\int_x^y (f'(t)-f'(x))dt.$$ Therefore $$\begin{align} f(y)-f(x)-f'(x)(y-x)&=\int_x^y (f'(t)-f'(x))dt \\ &\geq -\int_{\min(x,y)}^{\max(x,y)} |f'(t)-f'(x)|dt\\ &\geq -c\int_{\min(x,y)}^{\max(x,y)} |t-x|dt=-\frac{c}{2}(y-x)^2. \end{align}$$