IMO 2006 Shortlisted Problems

The proof goes by induction. For $n=1$ the formula yields $a_1=1/2$. Take $n\ge 1$, assume $a_1,\dots ,a_n>0$ and write the recurrence formula for $n$ and $n+1$, respectively as $$\sum _{k=0}^n\frac{a_k}{n-k+1}=0\text{ and }\sum _{k=0}^{n+1}\frac{a_k}{n-k+2}=0.$$ Subtraction yields $$\begin{array}{rl}0=(n+2)\sum _{k=0}^{n+1}& \frac{a_k}{n-k+2}-(n+1)\sum _{k=0}^n\frac{a_k}{n-k+1}\\ & =(n+2)a_{n+1}+\sum _{k=0}^n\left(\frac{n+2}{n-k+2}-\frac{n+1}{n-k+1}\right)a_k\end{array}$$ The coefficient of $a_0$ vanishes, so $$a_{n+1}=\frac{1}{n+2}\sum _{k=1}^n\left(\frac{n+1}{n-k+1}-\frac{n+2}{n-k+2}\right)a_k=\frac{1}{n+2}\sum _{k=1}^n\frac{k}{(n-k+1)(n-k+2)}{a}_k.$$ The coefficients of $a_1,\dots ,a_n$ are all positive. Therefore, $a_1,\dots ,a_n>0$ implies $a_{n+1}>0$.