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 $b_n = a_{n-1} + b_{n-1} + b_{n-2} > a_{n-1} + \frac{1}{3}(n-1)a_{n-1} + \frac{1}{3}(n-2)a_{n-2} = \frac{1}{3}n(a_{n-1} + a_{n-2}) + \frac{2}{3}(a_{n-1} - a_{n-2}) > \frac{1}{3}na_n$, where the last inequality holds as $a_{n-1} = a_{n-2} + a_{n-3} > a_{n-2}$.

By induction we get that $b_n > \frac{1}{3}na_n$ 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 $b_n = a_{n-1} + b_{n-1} + b_{n-2} < a_{n-1} + (n-1)a_{n-1} + (n-2)a_{n-2} = n(a_{n-1} + a_{n-2}) - 2a_{n-2} < na_n$.

By induction, $b_n < na_n$ holds for all $n \ge 3$.