Finding an upper limit on a continuous function I'm tasked with finding a smallest constant $c$ so that for a function $f\in C^4([0,1])$ and $x,y\in[0,1]$:
$$\left\vert\frac{f(x)-f(y)}{x-y}-\frac{f'(x)+f'(y)}{2}\right\vert\le c\vert x-y\vert ^2\max_\limits{\xi\in[0,1]}\vert f'''(\xi)\vert$$
Since this is an excercise I was given on the Topic of Taylor series, my first thought was making use of Taylor's theorem and somehow set the upper bound using the Lagrange remainder but I'm not quite sure that's the way to go. Alternatively I thought about using the fact, that $f(x)$ is lipschitz continuous. Any nudge in the right direction would be greatly appreciated. Thanks in advance.

For $x \ne y$, there exists $\xi_1, \xi_2$ between $x,y$ such that:
\begin{align}
f(x)&=f(y)+f'(y)(x-y)+\frac {f''(y)}2(x-y)^2+\frac{f'''(\xi_1)}6(x-y)^3\\
f'(x)&=f'(y)+f''(y)(x-y)+\frac {f'''(\xi_2)}2(x-y)^2\end{align}
Substituting:
\begin{align}&\quad\left\vert\frac{f(x)-f(y)}{x-y}-\frac{f'(x)+f'(y)}{2}\right\vert
\\&=\left\vert\frac{f'(y)(x-y)+\frac {f''(y)}2(x-y)^2+\frac{f'''(\xi_1)}6(x-y)^3}{x-y}-\frac{2f'(y)+f''(y)(x-y)+\frac {f'''(\xi_2)}2(x-y)^2}{2}\right\vert\\
\\&=\left\vert f'(y)+\frac {f''(y)}2(x-y)+\frac{f'''(\xi_1)}6(x-y)^2-f'(y)-\frac {f''(y)}2(x-y)-\frac {f'''(\xi_2)}4(x-y)^2\right\vert
\\&=\left\vert \frac{f'''(\xi_1)}6(x-y)^2-\frac {f'''(\xi_2)}4(x-y)^2\right\vert
\\&\le\left(\left\vert \frac{f'''(\xi_1)}6\right\vert+\left\vert\frac {f'''(\xi_2)}4\right\vert\right)(x-y)^2
\end{align}
