IMO Team Selection Contest

The definition of sequence $(b_n)$ indicates that $b_n-b_{n-2}=a_n-a_{n-2}+b_{n-1}$ for all $n\ge 3$. Therefore $b_n-b_{n-2}=a_{n-1}+b_{n-1}$, i.e., $b_n=a_{n-1}+b_{n-1}+b_{n-2}$.

Notice that the left-hand inequality $\frac{1}{3}<\frac{b_n}{n\cdot a_n}$ holds for $n=2$ and $n=3$ as $\frac{b_2}{2a_2}=\frac{a_2+b_1}{2a_2}=\frac{a_1+a_2}{2a_2}>\frac{a_2}{2a_2}=\frac{1}{2}>\frac{1}{3}$ and $\frac{b_3}{3a_3}=\frac{a_3+b_2}{3a_3}=\frac{2(a_1+a_2)}{3(a_1+a_2)}=\frac{2}{3}>\frac{1}{3}$.

Now assume that $n\ge 4$ and that the statement is true for $n-1$ and $n-2$. Then $b_n=a_{n-1}+b_{n-1}+b_{n-2}>a_{n-1}+\frac{1}{3}(n-1)a_{n-1}+\frac{1}{3}(n-2)a_{n-2}=\frac{1}{3}n(a_{n-1}+a_{n-2})+\frac{2}{3}(a_{n-1}-a_{n-2})>\frac{1}{3}n{a}_n$, where the last inequality holds as $a_{n-1}=a_{n-2}+a_{n-3}>a_{n-2}$.

By induction we get that $b_n>\frac{1}{3}n{a}_n$ for all $n\ge 2$.

Now notice that the right-hand inequality $\frac{b_n}{n\cdot a_n}<1$ holds for $n=3$ and $n=4$ as $\frac{b_3}{3a_3}=\frac{2}{3}<1$ and $\frac{b_4}{4a_4}=\frac{a_4+b_3+b_1}{4a_4}=\frac{a_4+a_3+a_2+a_1}{4a_4}<1$.

Let $n\ge 5$ and assume that the statement is valid for $n-1$ and $n-2$. Then $b_n=a_{n-1}+b_{n-1}+b_{n-2}<a_{n-1}+(n-1)a_{n-1}+(n-2)a_{n-2}=n(a_{n-1}+a_{n-2})-2a_{n-2}<n{a}_n$.

By induction, $b_n<n{a}_n$ holds for all $n\ge 3$.