Consider a sequence $\{x_n\}_{n=1}^\infty$ satisfying $x_{n+1} \leq a x_n + b \sqrt{x_n}+c$ with $x_1=0$, where $a\in (0,1)$, $b>0$, $c>0$, and $\Delta = b^2-4ac<0$. What is the best upper bound of $x_n$ we can get?
My attempt is to rewrite the recursive relation as $x_{n+1}\leq a(\sqrt{x_n}+\frac{b}{2a})^2 + \frac{4ac-b^2}{4a}$. Since $\sqrt{x^2+y^2} \leq x + y$ for any $x,y\geq0$, we have $\sqrt{x_{n+1}}\leq \sqrt{a}(\sqrt{x_n}+\frac{b}{2a}) + \sqrt{\frac{4ac-b2}{4a}}$. Then we can easily derive the upper bound of $\sqrt{x_n}$ and thus $x_n$.
Is there a better way to get an upper bound of $x_n$? I think the usage of $\sqrt{x^2+y^2}\leq x+y$ is too loose. Thanks!
We have $$\begin{align} &\iff x_{n+1}\leq a\left(\sqrt{x_n}+\frac{b}{2a}\right)^2 + \frac{4ac-b^2}{4a} \\ &\iff \underbrace{x_{n+1} + \frac{b^2}{4a^2}}_{\text{apply } x^2+y^2 \ge\frac{(x+y)^2}{2}} \leq a\left(\sqrt{x_n}+\frac{b}{2a}\right)^2 + \frac{4ac-b^2}{4a}+\frac{b^2}{4a^2} \\ &\Longrightarrow \frac{1}{2}\left(\sqrt{x_{n+1}}+\frac{b}{2a}\right)^2 \le a\left(\sqrt{x_n}+\frac{b}{2a}\right)^2 + \frac{4ac-b^2}{4a}+\frac{b^2}{4a^2}\\ &\Longrightarrow \left(\sqrt{x_{n+1}}+\frac{b}{2a}\right)^2 \le 2a\left(\sqrt{x_n}+\frac{b}{2a}\right)^2 + \frac{4ac-b^2}{2a}+\frac{b^2}{2a^2} \tag{1} \end{align}$$ Denote $y_n := \left(\sqrt{x_n}+\frac{b}{2a}\right)^2$ and $d:=\frac{4ac-b^2}{2a}+\frac{b^2}{2a^2} $, then $$(1) \iff y_{n+1} \le 2a\cdot y_n + d$$ $$\Longrightarrow y_{n+1} \le 2a\cdot y_n + d \le (2a)^2 \cdot y_{n-1}+d(1+2a) \le...\le (2a)^{n}y_1+d\frac{1-(2a)^n}{1-2a} \tag{2}$$
So, from $(2)$, a tighter bound for $x_n$ would be $$\begin{align}x_n &=\left(\sqrt{y_n}-\frac{b}{2a}\right)^2 \\ &\le \color{red}{\left(\sqrt{(2a)^{n-1}\left(\sqrt{x_1}+\frac{b}{2a}\right)^2+\left(\frac{4ac-b^2}{2a}+\frac{b^2}{2a^2} \right)\frac{1-(2a)^{n-1}}{1-2a}}-\frac{b}{2a}\right)^2}\end{align}$$