From concavity to second derivative Let $f(x)$ be continuous and differentiable on $\mathbb{R}$. I want to show that if $f(x)$ is concave, then $f''(x)\le0$ $\forall\space x\in\mathbb{R}$
Proof:
Let $x,y$ be fixed in $\mathbb{R}$ such that $y>x$
Without loss of generality, since $f$ is concave, then $f'(x)\ge\cfrac{f(y)-f(x)}{y-x}$ by the calculus criterion.
By the mean value theorem, $\exists\space c\in[x,y]$ such that $f'(c)=\cfrac{f(y)-f(x)}{y-x}$
So I now have: $f'(x)\ge f'(c)$
Since $x$ was arbitrarily chosen, then we have shown $a\ge b\implies f'(b)\ge f'(a)$
Since $y>x$, then we have $f'(x)\ge f'(y)$ and this is true for all $x,y\in\mathbb{R}$
So $f''(x)\le 0$
Is my approach correct?
 A: Aside from the incorrect reasoning after you apply the MVT, you need to assume that $f$ is twice differentiable, since not all concave functions are.
For $x_1 < x < y < x_2$ and $f$ concave you can show that
$$\tag{1}\frac{f(x) - f(x_1)}{x-x_1} \geqslant \frac{f(x_2) - f(y)}{x_2 -y}$$
Taking limits as $x \to x_1+$ and $y \to x_2-$ we get
$$f'(x_1) \geqslant f'(x_2),$$
showing that $f'$ is decreasing and $f''(x) \leqslant 0$.
Proof of (1)
If $x_1 < x < y$ then 
$$x = \frac{y - x}{y - x_1}x_1 + \frac{x-x_1}{y-x_1}y$$
By concavity
$$f(x) \geqslant \frac{y - x}{y - x_1}f(x_1) + \frac{x-x_1}{y-x_1}f(y)$$
Hence,
$$f(x) - f(x_1) \geqslant \left(\frac{y - x}{y - x_1}-1\right)f(x_1) + \frac{x-x_1}{y-x_1}f(y) \\ = \frac{x-x_1}{y- x_1}(f(y) - f(x_1)),$$
and, thus,
$$\tag{2}\frac{f(x) - f(x_1)}{x-x_1} \geqslant \frac{f(y) - f(x_1)}{y-x_1}$$
In the same way for $x_1 < y < x_2$ we can show
$$\tag{3}\frac{f(y) - f(x_1)}{y-x_1} \geqslant \frac{f(x_2) - f(y)}{x_2-y}$$
Therefore, from (2) and (3) we have
$$\frac{f(x) - f(x_1)}{x-x_1} \geqslant \frac{f(x_2) - f(y)}{x_2-y}$$
