Given $$a_{n+1} = \sqrt{\dfrac{ab^2+a_n^2}{a+1}}\,\,\forall n$$ Where $a>0, 0< a_1< b, a=a_1$
I obtained the expression for terms $a_2^2, a_3^2, a_4^2,...,a_n^2$ and do the summation of $a_i^2-a_{i-1}^2$ but that results in $a_n^2-a_2^2= \dfrac{a_{n-1}^2-a_1^2}{a+1}$, which I have no clue what to do with it further. Any little trick or suggestions would be appreciated.
Note first that $$ a_2^2-a_1^2 = \frac{ab^2+a_1^2}{a+1}-a_1^2 = \frac{ab^2+a^2}{a+1}-\frac{a^2(a+1)}{a+1}= \frac{a(b^2-a^2)}{a+1}. $$ We get that $a_2^2-a_1^2$, as the right hand side is positive, and so $a_2>a_1$ (these are square roots, hence positive).
Also, note that $$ a_{n+1}^2 - a_n^2 = \frac{ab^2+a_n^2}{a+1}-\frac{ab^2+a_{n-1}^2}{a+1} = \frac{a_n^2-a_{n-1}^2}{a+1}. $$ This way, $a_n>a_{n-1}$ implies $a_{n+1}>a_n$, and so by induction the sequence is increasing.