67th Romanian Mathematical Olympiad

Let $F:\mathbb{R}\to \mathbb{R}$, $F(x)={\int }_a^{x}f(t)dt$. If $f$ is continuous at $a$, then $F$ is differentiable at $a$ and $F^{\prime }(a)=f(a)$. For every positive integer $n$, there exists a positive real number ${\delta }_n$ such that $|F(x)/(x-a)-f(a)|\le 1/(2n)$ for all $x$ in the ${\delta }_n$-neighbourhood of $a$. Let $a_n={\delta }_n/2$, $n=1,2,3,\dots$, to get $F(a+a_n)/a_n-f(a)\le 1/(2n)$ and $f(a)-F(a-a_n)/(-a_n)\le 1/(2n)$, so $F(a+a_n)+F(a-a_n)\le a_n/n$, by addition of the two.

Conversely, since $f$ is increasing, so is $g:\mathbb{R}\setminus \{a\}\to \mathbb{R}$, $g(x)=F(x)/(x-a)$. Indeed, let $x,y\in \mathbb{R}\setminus \{a\}$, $a<x<y$; the cases $x<y<a$ and $x<a<y$ are dealt with similarly. Since $(x-a)F(y)=(x-a){\int }_a^{x}f(t)dt+(x-a){\int }_x^{y}f(t)dt\ge (x-a){\int }_a^{x}f(t)dt+(x-a)(y-x)f(x)\ge (x-a){\int }_a^{x}f(t)dt+(y-x){\int }_a^{x}f(t)dt=(y-a)F(x)$, it follows that $g(y)\ge g(x)$.

For every positive integer $n$, choose $x_n$ in the open interval $(0,a_n)$ so that $x_n\stackrel{n\to \infty }{\to }0$. Since $g$ is increasing, it follows that $g(a+x_n)-g(a-x_n)\le g(a+a_n)-g(a-a_n)\le 1/n$, $n=1,2,3,\dots$, so $g(a+0)\le g(a-0)$. On the other hand, $g(a-0)\le g(a+0)$ for $g$ is increasing, so $g(a+0)=g(a-0)$ is finite, and $F$ is differentiable at $a$. The conclusion follows.