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The answer is no. E.g., let $\alpha = 2,\beta = 1,f_0= 0,f_1= 2,f_2= 4,f_3= 11$. Then $f_{n+1} \leq \alpha f_n + \beta f_{n-1}$ for $n=1,2$, whereas $$ f_3 = 11 \not\leq 10= \left(\alpha ^2+\beta \right)f_1 +\alpha \beta f_0=[x^3] \dfrac{f_0 + xf_1 - \alpha x f_0}{1-\alpha x - \beta x^2}. $$

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

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 $$ f_3 = \frac{49}{128} \not\leq 0= \left(\alpha ^2+\beta \right)f_1 +\alpha \beta f_0=[x^3] \dfrac{f_0 + xf_1 - \alpha x f_0}{1-\alpha x - \beta x^2}. $$

The answer is no. E.g., let $\alpha = 2,\beta = 1,f_0= 0,f_1= 2,f_2= 4,f_3= 11$. Then $f_{n+1} \leq \alpha f_n + \beta f_{n-1}$ for $n=1,2$, whereas $$ f_3 = 11 \not\leq 10= \left(\alpha ^2+\beta \right)f_1 +\alpha \beta f_0=[x^3] \dfrac{f_0 + xf_1 - \alpha x f_0}{1-\alpha x - \beta x^2}. $$

The answer is no. E.g., let $f_0=0$, $f_1=1$, $\alpha=1$, $\beta=0$, $F(x)\equiv x/2+3x^2/2$, $I=(0,1)$. Then \begin{equation} F(x)=x/2+3x^2/2\le x+x^2+\dots= \dfrac{x}{1-x}=\dfrac{f_0 + xf_1 - \alpha x f_0}{1-\alpha x - \beta x^2} \end{equation} for $x\in I$, whereas $3/2\not\le1$.

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 $$ f_3 = \frac{49}{128} \not\leq 0= \left(\alpha ^2+\beta \right)f_1 +\alpha \beta f_0=[x^3] \dfrac{f_0 + xf_1 - \alpha x f_0}{1-\alpha x - \beta x^2}. $$