TITLE: I conjecture inequalities $\sum_{k=1}^{n}\{kx\}\le\frac{n}{2}x$
QUESTION [13 upvotes]: I conjecture the following inequality:

For $x > 1$, and $n$ a positive integer,
  $$\sum_{k=1}^{n}\{kx\}\le\dfrac{n}{2}x.$$

For $n=1$, the inequality becomes
$$\{x\}\le\dfrac{x}{2}\Longleftrightarrow \{x\}\le [x];$$
and, for $n = 2$, it becomes
$$\{x\}+\{2x\}\le x\Longleftrightarrow \{2x\}\le [x],$$
which is obvious since $[x]\ge 1>\{2x\}$.
If $x\ge 2$, the inequality is obvious, since
$$\dfrac{n}{2}x\ge n\ge\sum_{k=1}^{n}\{kx\}.$$
However, I cannot prove it for $1 < x < 2$.

REPLY [7 votes]: This is a proof smilar to Will Sawin's but with no induction.
Set $y=x-1$. We need to prove that the average of $\{y\},\dots,\{ny\}$ is at most $x/2$. The numbers $y,2y,\dots,ny$ split into contiguous groups, each group with the same integer part. It suffices to show that the average of the fractional parts for each group is at most $x/2$. Those fractional parts form an arithmetical sequence, so their average is half the sum of the first and the last term. As the fractional part of the first term in the group is at most $y$, this average is bounded by $\frac{y+1}2=x/2$, as desired.