Argentine National Olympiad 2016

We show that $S(nk)<S(n)+S(k)$. Cancel the summand of $S(k)$ on both sides of this inequality, then add $\frac{1}{2}$ to both sides and rearrange. This yields the equivalent inequality $$\frac{1}{k+1}+\frac{1}{k+2}+\cdots +\frac{1}{nk}+\frac{1}{2}<1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n}.(\ast )$$ Divide the first $(n-1)k$ numbers on the left hand side into $n-1$ sums of $k$ fractions with consecutive denominators: $A_1=\frac{1}{k+1}+\cdots +\frac{1}{2k}$, $A_2=\frac{1}{2k+1}+\cdots +\frac{1}{3k}$, ..., $A_{n-1}=\frac{1}{(n-1)k+1}+\cdots +\frac{1}{nk}$. Then compare $A_j$ with $\frac{1}{j}$ for $j=1,\dots ,n-1$. Denoting $d_j=\frac{1}{j}-A_j=\frac{1}{j}-\frac{1}{jk+1}-\frac{1}{jk+2}-\cdots -\frac{1}{jk+k}$ we have $$d_j=\left(\frac{1}{jk}-\frac{1}{jk+1}\right)+\left(\frac{1}{jk}-\frac{1}{jk+2}\right)+\cdots +\left(\frac{1}{jk}-\frac{1}{jk+k}\right)=$$ $$=\frac{1}{jk(jk+1)}+\frac{2}{jk(jk+2)}+\cdots +\frac{k}{jk(jk+k)}>$$

$$>\frac{1+2+...+k}{jk(jk+k)}=\frac{k+1}{2k}\cdot \frac{1}{j(j+1)}>\frac{1}{2j}-\frac{1}{2(j+1)}$$ It follows that $$\left(1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n}\right)-\left(\frac{1}{k+1}+\frac{1}{k+2}+\cdots +\frac{1}{nk}\right)=d_1+\cdots +d_{n-1}+\frac{1}{n}>\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{6}\right)+\cdots +\left(\frac{1}{2n-2}-\frac{1}{2n}\right)+\frac{1}{n}=\frac{1}{2}+\frac{1}{2n}>\frac{1}{2}$$ This proves (*), hence $S(nk)<S(n)+S(k)$ holds true.