THE 68th NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS

First solution. The required maximum is $n/2$ and is achieved if and only if $a_k=2^{k-2}{a}_1$, $k=2,\dots ,n$, and $a_1$ is any positive real number. To prove this, let $A_k=a_1+\cdots +a_k$, $k=0,\dots ,n-1$, where empty sums are zero, and refer to the condition in the statement, $A_k\le a_{k+1}$, $k=0,\dots ,n-1$, to write $$\begin{array}{rl}\sum _{k=1}^{n-1}\frac{a_k}{a_{k+1}}& =\sum _{k=1}^{n-1}\frac{A_k-A_{k-1}}{a_{k+1}}=\sum _{k=1}^{n-2}{A}_k\left(\frac{1}{a_{k+1}}-\frac{1}{a_{k+2}}\right)+\frac{A_{n-1}}{a_n}\\ & \le \sum _{k=1}^{n-2}{a}_{k+1}\left(\frac{1}{a_{k+1}}-\frac{1}{a_{k+2}}\right)+1=\sum _{k=1}^{n-2}\left(1-\frac{a_{k+1}}{a_{k+2}}\right)+1\\ & =n-1+\frac{a_1}{a_2}-\sum _{k=1}^{n-1}\frac{a_k}{a_{k+1}}\le n-\sum _{k=1}^{n-1}\frac{a_k}{a_{k+1}}.\end{array}$$ Consequently, ${\sum }_{k=1}^{n-1}{a}_k/a_{k+1}\le n/2$, and equality holds if and only if $A_k=a_{k+1}$, $k=1,\dots ,n-1$, which is clearly the case if and only if $a_k=2^{k-2}{a}_1$, $k=2,\dots ,n$, and $a_1$ is any positive real number.

Second solution. We now show by induction on $n\ge 2$ that $s_n(a_1,\dots ,a_n)=a_1/a_2+a_2/a_3+\cdots +a_{n-1}/a_n\le n/2$ for all positive real numbers $a_1,\dots ,a_n$ such that $a_k\ge a_1+\cdots +a_{k-1}$, $k=2,\dots ,n$, and equality holds if and only if $a_k=2^{k-2}{a}_1$, $k=2,\dots ,n$, and $a_1$ is any positive real number. Clearly, $s_2(a_1,a_2)\le 1$ if $0<a_1\le a_2$, and $s_3(a_1,a_2,a_3)\le s_3(a_1,a_2,a_1+a_2)\le 3/2$ if $0<a_1\le a_2$ and $a_1+a_2\le a_3$; in both cases, equality holds if and only if the $a_k$ are as stated. Now let $n\ge 4$, let $a_1,\dots ,a_n$ be positive real numbers such that $a_k\ge a_1+\cdots +a_{k-1}$, $k=2,\dots ,n$, and write $$\begin{array}{rl}{s}_n(a_1,\dots ,a_n)& \le s_n(a_1,\dots ,a_{n-1},a_1+\cdots +a_{n-1})\\ & =s_{n-2}(a_1,\dots ,a_{n-2})+\frac{a_{n-2}}{a_{n-1}}+\frac{a_{n-1}}{a_1+\cdots +a_{n-1}}.\end{array}$$ If we show that $$\frac{a_{n-2}}{a_{n-1}}+\frac{a_{n-1}}{a_1+\cdots +a_{n-1}}\le \frac{a_{n-2}}{a_1+\cdots +a_{n-2}}+\frac{1}{2},(\ast )$$ then $s_n(a_1,\dots ,a_n)\le s_{n-1}(a_1,\dots ,a_{n-2},a_1+\cdots +a_{n-2})+1/2\le (n-1)/2+1/2=n/2$, by the induction hypothesis; the cases of equality also follow from the induction hypothesis. Finally, to establish (*), simply notice that the numerator of the right-hand member and the left-hand member is the product of $a_{n-1}-a_1-\cdots -a_{n-2}\ge 0$ and $a_{n-1}(a_{n-2}-a_1-\cdots -a_{n-3})+2a_{n-2}(a_1+\cdots +a_{n-2})\ge 2a_{n-2}(a_1+\cdots +a_{n-2})>0$.