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The answer is no. E.g., let $\alpha=0,\beta=\frac{7}{8},f_0=1,f_1=0,f_2=\frac{7}{16},f_3=\frac{49}{128}$. Then $f_{n+1} \leq \alpha f_n + \beta f_{n-1}$ for $n=1,2$, whereas $$ f_3 = \frac{49}{128} \not\leq 0= \left(\alpha ^2+\beta \right)f_1 +\alpha \beta f_0=[x^3] \dfrac{f_0 + xf_1 - \alpha x f_0}{1-\alpha x - \beta x^2}. $$