$f \in C(\mathbb R)$ such that $f\Big(x+\dfrac 1n \Big) > f\Big(x-\dfrac 1n \Big) , \forall x \in \mathbb R , n \in \mathbb N$ ; is $f$ increasing? Let $f:\mathbb R \to \mathbb R$ be a continuous function such that $f\Big(x+\dfrac 1n \Big) > f\Big(x-\dfrac 1n \Big) , \forall x \in \mathbb R , n \in \mathbb N$ , then is it true that $f$ is increasing ? 
 A: Put $x = y+\frac 1 n$ for arbitrary $y$; so $f(y) < f(y + \frac 2 n)$ for all $y\in \mathbb R$ and $n \in \mathbb N^*$. Put $n=2k$, this means $f(y) < f(y+\frac 1 k)$ for all $y, k$. Now: repeating, $f(y) < f(y+\frac p k)$ for all positive integers $p$. Then for any $z>y$ choose a sequence of positive rational numbers that converges to $z-y$; this will show $f(y) <= f(z)$, and in fact the inequality is strict because $f(y)$ is already $<f(y+\frac 1 n)$ for some $n > \frac 1 {z-y}$, and you can apply the non-strict inequality to $y+\frac 1 n$ and $z$.
A: Yes. First to smooth out some irrelevant details: The given inequality is the same as $$f(x+2/n)>f(x)\quad(x\in\Bbb R),$$which implies $$f(x+1/n)>f(x),$$and hence $$f(x+r)>f(x)\quad(r>0,r\in\Bbb Q).$$
Now say $a<b$. Choose a sequence of rationals $r_n\in(0,b-a)$ such that $r_n$ increases to $ b-a$. Since $f$ is continuous we have $$f(a)<f(a+r_1)\le f(a+r_n)\to f(b),$$so $$f(b)\ge f(a+r_1)>f(a).$$
You might note that it's false if $f$ is not continuous, for example $$f(x)=\begin{cases}
x,&(x\in\Bbb Q),
\\x+1,&(x\notin\Bbb Q).
\end{cases}$$
