Just for simplicity, define $b_k = a_k+a_{k-1}$. Our condition transforms into $b_k * b_{k+1} = b_k - b_{k+1}$ So: $ b_k * b_{k+1} - b_k + b_{k+1}=0$ $=> (b_k +1)(b_{k+1}-1) = -1$ $=> (a_{k-1}+a_k+1)(a_k+a_{k+1}-1) = -1$ . Call that the new ecuation. Since the $a_i$ are positive reals we have that $a_k+a_{k+1} - 1 < 0$ $=> a_k+a_{k+1} < 1$. Therefore, for $k=1$ we get $a_2<a_1+a_2<1= \frac{1}{2-1}$. Also, since $a_1+a_2<1$ we have $a_1+a_2+1<2$. By plugging $k=2$ in the new ecuation we get $a_1+a_2+1= \frac{-1}{a_2+a_3-1}$ and therefore $\frac{-1}{a_2+a_3-1} <2$ and considering that $a_2+a_3-1$ is negative we solve the inequality and we get $\frac{1}{2}> a_2+a_3>a_3$ which is what we wanted to prove for n=3. We will show by induction over $n$ that $a_{n-1}+a_{n} < \frac{1}{n-1}$ . From this the result will follow since $a_{n}<a_{n-1}+a_{n}$. The base cases $n=2$ and $n=3$ have already been proven. Now suppose that the inequality is true for $n-1$. We prove that it is true for $n$. By the induction hypothesis we have: $a_{n-2}+a_{n-1} < \frac{1}{n-2}$ and therefore $a_{n-2}+a_{n-1} +1< \frac{1+n-2}{n-2}= \frac{n-1}{n-2}$ By plugging $n-1$ in the new ecuation we have $(a_{n-2}+a_{n-1}+1)(a_{n-1}+a_n-1)=-1$, hence $\frac{-1}{a_{n-1}+a_n-1}< \frac{n-1}{n-2}$ $=> \frac{-n+2}{n-1}> a_{n-1}+a_n - 1$ $=> a_{n-1}+a_n < \frac{-n+2}{n-1} + 1 = \frac{-n+2+n-1}{n-1}= \frac{1}{n-1}$ and we are done. *EDIT: Keeping the notation $b_k = a_k+a_{k-1}$ would have made things cleaner.

Better $c_k=\frac{1}{b_k}$. Then $c_{k+1}=c_k+1$. Therefore $a_n<b_n=\frac{1}{c_n}=\frac{1}{n-1+c_1}<\frac{1}{n-1}.$

Rust wrote: Better $c_k=\frac{1}{b_k}$. Then $c_{k+1}=c_k+1$. Therefore $a_n<b_n=\frac{1}{c_n}=\frac{1}{n-1+c_1}<\frac{1}{n-1}.$ Solutions are never clean the first time you write them

Similar problem EGMO 2022 P4.