75th NMO Selection Tests

Clearly, the $s_n$ and the ${\sigma }_n$ both form strictly increasing sequences. We will construct a sequence of positive integers $n_1<n_2<\cdots <n_k<\dots$ such that ${\sigma }_{n_k}>k{a}_1/(1+a_1)$ for all $k\ge 2$. The conclusion then follows at once. Let $n_1$ be any positive integer, then use unboundedness of the $s_n$ to choose $n_k$ recursively so that $s_{n_k}>s_{n_{k-1}}+a_1$, $k\ge 2$. The conclusion is a consequence of the following inequality: If $m\ge 2$ and $x_1,x_2,\dots ,x_m$ are positive real numbers, then $$\frac{x_1}{1+x_1}+\frac{x_2}{1+x_2}+\cdots +\frac{x_m}{1+x_m}>\frac{x_1+x_2+\cdots +x_m}{1+x_1+x_2+\cdots +x_m}.(\ast )$$ Assume $(\ast )$ for the moment to write $$\begin{array}{rl}{\sigma }_{n_k}-{\sigma }_{n_{k-1}}& =\frac{a_{n_{k-1}+1}}{1+a_{n_{k-1}+1}}+\cdots +\frac{a_{n_k}}{1+a_{n_k}}\\ & >\frac{a_{n_{k-1}+1}+\cdots +a_{n_k}}{1+a_{n_{k-1}+1}+\cdots +a_{n_k}}=\frac{s_{n_k}-s_{n_{k-1}}}{1+s_{n_k}-s_{n_{k-1}}}>\frac{a_1}{1+a_1},k\ge 2,\end{array}$$ so ${\sigma }_{n_k}-{\sigma }_{n_1}>(k-1)a_1/(1+a_1)$. As ${\sigma }_{n_1}\ge {\sigma }_1=a_1/(1+a_1)$, it follows that ${\sigma }_{n_k}>k{a}_1/(1+a_1)$ for all $k\ge 2$, as desired.

Finally, we prove $(\ast )$ in two different ways.

1st Proof. Write $$\begin{array}{rl}\sum _{k=1}^m\frac{x_k}{1+x_k}& =\sum _{k=1}^m\frac{x_k^2}{x_k+x_k^2}\ge \frac{{\left({\sum }_{k=1}^m{x}_k\right)}^2}{{\sum }_{k=1}^m{x}_k+{\sum }_{k=1}^m{x}_k^2},\text{by Cauchy-Schwarz}\\ & >\frac{{\left({\sum }_{k=1}^m{x}_k\right)}^2}{{\sum }_{k=1}^m{x}_k+{\left({\sum }_{k=1}^m{x}_k\right)}^2}=\frac{{\sum }_{k=1}^m{x}_k}{1+{\sum }_{k=1}^m{x}_k}.\end{array}$$

2nd Proof. Induct on $m$. The base case, $m=2$, is a routine check. For $m\ge 3$, $$ $$\begin{array}{rl}\sum _{k=1}^m\frac{x_k}{1+x_k}& =\sum _{k=1}^{m-2}\frac{x_k}{1+x_k}+\left(\frac{x_{m-1}}{1+x_{m-1}}+\frac{x_m}{1+x_m}\right)\\ & >\sum _{k=1}^{m-2}\frac{x_k}{1+x_k}+\frac{x_{m-1}+x_m}{1+x_{m-1}+x_m}>\frac{{\sum }_{k=1}^m{x}_k}{1+{\sum }_{k=1}^m{x}_k}.\end{array}$$