any ideas?

Let $c_n$ define the sequence equation that is given for even n>1;similarly define $d_n$ to be sequence equation of odd n>1. Take $c_n$ and $c_{n-2}$ and subtract them. This gives $a_{n-1} + b_{n-1} + b_{n-2} = b_{n}$....(1) Now we get Into the way of induction. Base case:[n=3] $b_3 = a_3 + b_2 = a_3 + a_2 + b_1 <a_3 + a_3+a_3$, Which proves the case when n =3. We can also show n=4 works. Now say $b_m < m.a_m$ for m=3,4,5,...,n-1 From (1),$b_n<a_{n-1} + (n-1)a_{n-1} +(n-2)a_{n-2}$ Now we take out the weapon: $a_n = a_{n-1} + a_{n-2}$ Using it we get $b_n<n.(a_n-a_{n-2}) + (n-2)a_{n-2}$ So $b_n<n.a_n - 2.a_{n-2} < n.a_n$ We can prove $b_n<n.a_n$for even n in the almost same way. I will post the solution the solution for left side as well (I am actually doing this sum)

Now for the left side of the inequality:) Assume for the sake of strong induction, $b_m > \frac{m}{3}a_m$ for m=3,4,...,n-1 ; n is even. From equation (1), $b_n>a_{n-1} + \frac{n-1}{3}a_{n-1}+ \frac{n-2}{3}a_{n-2}$ So $b_n > \frac{n}{3}a_n + \frac{2}{3} (a_n-2a_{n-2})$ $a_n> 2a_{n-2}$ So completes the proof for even n. Almost same for odd n. Sum's done!!!