If $f : [a,b]\to\Bbb R$ is continuous, are there $x_1,x_2\in (a,b)$ such that $\tfrac{f(b)-f(a)}{b-a} = \tfrac{f(x_1)-f(x_2)}{x_1-x_2}$? I just thought about the mean value theorem and wondered whether the following statement is true:

If $f : [a,b]\to\Bbb R$ is continuous, then there are $x_1,x_2\in (a,b)$
  such that $\tfrac{f(b)-f(a)}{b-a} = \tfrac{f(x_1)-f(x_2)}{x_1-x_2}$.

One way to look at it is to consider the function $F : \{(x,y) : x,y\in[a,b],\,y>x\}\to\Bbb R$, defined by $F(x,y) = \tfrac{f(x)-f(y)}{x-y}$. If there don't exist such $x_1,x_2$, then $(a,b)$ is a maximum of a minimum of $F$. But I don't know what to conclude from that. Does anybody have an idea?

I think this might be a rather geometric proof. 
Consider a continuous function $f\colon [a,b]\to \mathbb{R}$. Let $\ell$ be the line that goes through the points $(a,f(a))$ and $(b,f(b))$.
If the graph of $f$ is equal to $\ell$ (or a segment of it, to be more precise), then there is nothing to prove. Note that in this case
$$f(x) = \frac{f(b)-f(a)}{b-a}(x-a) +f(a).$$
If the graph of $f$ is not equal to the line $\ell$, then there is one point $(x,f(x))$ on the graph of $f$ that does not lie on the line segment.  
Consider a line $\tilde \ell$ that is parallel to the line $\ell$ and lies "between" the point $(x,f(x))$ and the line $\ell$ (i.e. $\tilde \ell$ seperates the point $(x,f(x))$ and the line $\ell$). Now, since $f$ is continuous and the interval $[a,b]$ is connected, the graph of $f$ is also connected. Therefore, the graph of $f$ must intersect $\tilde \ell$ at a point $(x_1,f(x_1))$ with $x_1 \in (a,x)$ and also at another point $(x_2,f(x_2))$ with $x_2 \in (x,b)$. Since these two points lie on $\tilde \ell$, we have
$$ \frac{f(b)-f(a)}{b-a} = \frac{f(x_2)-f(x_1)}{x_2-x_1}.$$
