Proof of Cauchy's mean value theorem and Lagrange's mean value theorem that does not depend on Rolle's theorem Can you give one elegant proof of Cauchy's Mean Value Theorem and one of Lagrange's Mean Value Theorem (*) which do not depend on Rolle's theorem?
$_{\text{(*) A special case of Cauchy's.}}$
 A: If you have somehow proved Cauchy's theorem, you have as a bonus also Rolle's and Lagrange's. Conversely, Rolle's theorem easily implies Cauchy's, via an affine transformation. Thus the three theorems are equivalent to one another. Proving one is the same as proving all three of them.
Rolle's theorem for $f$ continuous on $[a,b]$ and differentiable in $(a,b)$  states the existence of an internal critical point for the function $f$. I don't think there's a way for proving such existence without using the fact that a continuous function on a closed and bounded interval has maximum and minimum.
Even if we assume the derivative is continuous on $(a,b)$ we cannot appeal to the fact that $f$ is monotonic if the derivative doesn't vanish, because this follows from Lagrange's theorem.
A: You can give an alternative proof of the mean value theorem of Lagrange by means of the fundamental theorem of calculus. This is slightly less efficient, though, because you need to assume that the function is differentiable with a continuous derivative.
The proof is very easy. Let $f\colon [a, b]\to \mathbb R$ be differentiable, with $f'\colon (a, b)\to \mathbb R$ continuous. Then, for all $a<x_1<x_2<b$, 
$$\tag{1}
f(x_2)-f(x_1)=\int_{x_1}^{x_2} f'(y)\, dy, $$ 
and by the mean value theorem for integrals, there exists $\xi\in (x_1, x_2)$ such that
$$ \tag{2}
\int_{x_1}^{x_2} f'(y)\, dy=f'(\xi)(x_2-x_1). $$
$\Box$
Remark. Egreg points out that the proof of the formula of Lagrange depends on the theorem of Weierstrass, which states that a continuous function attains maximum and minimum on a compact interval. The proof presented here is no exception. Indeed, while (1) holds in full generality, even for functions $f\colon [a, b]\to X$, where $X$ is a Banach space, the mean value formula (2) is proved by using the theorem of Weierstrass. 
This is unavoidable, as (2) fails in cases when Weierstrass is not available. For example, (2) fails for the function $f\colon [0, 1]\to \mathbb C$ defined by $$ f(\theta)= e^{2\pi i \theta}.$$
A: Here is a proof of Lagrange 's Mean Value Theorem without use of Rolle 's Theorem.
Theorem. Let a function $f:[a,b]\to \Bbb R$ continuous in $[a,b]$ and differentiable in $(a,b)$. Then there exists a $ξ\in(a,b)$, such that $f'(ξ)=\frac {f(b)-f(a)}{b-a}$.
Proof. Set $ω:=\frac {f(b)-f(a)}{b-a}$. If there exist $x_1, x_2\in (a,b)$, such that $f'(x_1)<ω<f'(x_2)$, then the Darboux 's Intermediate Value Theorem completes the proof. Suppose that $\forall x\in (a,b): f'(x)<ω$ (the reasoning is similar for $f'(x)>ω$). The function $g(x):=f(x)-ωx, x\in [a,b]$ maintains the properties of $f$, $g(a)=g(b) (*)$ and $\forall x\in (a,b):g'(x)=f'(x)-ω<0$, which means that $g$ is decreasing in $(a,b)$. By this fact and the continuity of $g$ in $[a,b]$ we get $g(a)>g(b)$ (because $\forall ε\in (0,\frac{b-a}{3}):a+ε<a+\frac{b-a}{3}<b-\frac{b-a}{3}<b-ε \Rightarrow g(a+ε)>g(a+\frac{b-a}{3})>g(b-\frac{b-a}{3})>g(b-ε)\Rightarrow\lim_{ε\to 0^+}g(a+ε)\geq g(a+\frac{b-a}{3})>g(b-\frac{b-a}{3}) \geq lim _{ε \to 0^+}g(b-ε) \Rightarrow g(a)>g(b))$, which contradicts to $(*)$. q.e.d.
