Cono Sur Mathematical Olympiad

We will actually prove a stronger statement: if for some $k$ it is true that $S_k<S_{k+1}$, then $S_j<S_{j+1}$ for all $j\ge k$ (i.e., the sequence is strictly increasing starting from that point).

For our proof, we will use the following key observation. For any positive real number $x$ and positive integer $m$, the inequality $x^{m+2}-x^{m+1}\ge x^{m+1}-x^m$ holds. Indeed, dividing everything by $x^m$ we obtain the equivalent inequality $x^2-x\ge x-1$, that is $x^2-2x+1\ge 0$, which is true because the left hand side equals $(x-1{)}^2$.

Now it suffices to see that by letting $m$ fixed and adding up all these inequalities for each number $x_i$, we conclude that $S_{m+2}-S_{m+1}\ge S_{m+1}-S_m$. In particular, if for some $k$ it happens that $S_{k+1}>S_k$, then for all $j\ge k$ we have that $$S_{j+1}-S_j\ge S_j-S_{j-1}\ge \cdots \ge S_{k+1}-S_k>0,$$ as claimed.

To construct a counterexample for part (b), we can take $n=2$, $x_1=1.1$ and $x_2=0.5$. For these numbers, $S_1=1.6$ and $S_2=1.21+0.25=1.46<S_1$. However, the sequence is not strictly decreasing, because for a large enough $n$ we have that $S_n>1.1^n>1.6=S_1$.