Solving the recurrence relation $a_n=\frac{a_{n-1}^2+a_{n-2}^2}{a_{n-1}+a_{n-2}}$

The sequence of ratios $r_n=a_{n+1}/a_n$ is positive and such that $r_{n+1}-1=-\frac{r_n-1}{r_n(1+r_n)}$, hence $r_n-1$ is alternatively positive and negative. Since $r_1=a_2/a_1=2\gt1$, one gets $a_{2n}\gt a_{2n+1}$ and $a_{2n-1}\lt a_{2n}$ for every $n\geqslant1$. Likewise, one can show that $(a_{2n})_{n\geqslant1}$ is decreasing while $(a_{2n-1})_{n\geqslant1}$ is increasing. Finally, $r_n\to1$ hence $(a_{2n})_{n\geqslant1}$ and $(a_{2n-1})_{n\geqslant1}$ converge to the same limit $\ell$ and this limit is such that $a_{2n-1}\lt\ell\lt a_{2n}$ for every $n\geqslant1$ (this holds for every initial conditions $0\lt a_1\lt a_2$, otherwise one should exchange the odd numbered and the even numbered terms).

Until somebody finds a way to compute analytically the limit $\ell$ of $(a_n)_{n\geqslant1}$, one should rely on numerical approximations, based on the inequalities $a_{2n-1}\lt\ell\lt a_{2n}$ mentioned above. These show in particular that $\ell\gt a_3=\frac53$, hence $\ell$ is neither $\frac32$ the arithmetic mean of $a_1=1$ and $a_2=2$, nor $\frac43$ its harmonic mean, nor $\sqrt2$ its geometric mean.