Decreasing function depending in parameters Consider :


*

*$\alpha$ , $\lambda$ and $\rho$ $\in \mathbb{R}$ 

*$\alpha$ $\geq$ - $\frac{1}{2}$

*$a=\alpha$ + $\frac{1}{2}$ and $b=\frac{\lambda^2}{2(\alpha+1)}$

*a function $f(x)\in \mathbb{R}$   ,  $x>0$

*$f$ is decreasing to $2\rho$ 
ie:  $f$ is decreasing and $$\lim_{x\rightarrow\infty} f(x) = 2\rho\;
 \forall x > 0.$$
Consider a function $H:$ $$H(x) = \frac{a}{x} - \frac{1}{2}f(x) - b\ , \forall x > 0.$$ 
I need to know:


*

*in which cases $$\exists\ c \in \mathbb{R}\  : H(x) < c\ , \forall x > 0.$$

*in these cases what are the values of $c$ ?


Thanks.
Update
$$f(x) = \frac{2 \alpha + 1}{x} + \beta(x)\ , \forall x > 0.$$
where $\beta$ is a function and $\beta(0)=0$
 A: We see that
$$H(x)\lt c$$
is equivalent to
$$f(x)\gt \frac{2\alpha +1}{x} - \frac{\lambda^2}{\alpha+1}-2c\tag1$$


*

*Case 1 : If $\alpha=-\frac 12$, then $(1)$ becomes
$$f(x)\gt - 2\lambda^2-2c\tag2$$If $f(x)\gt 2\rho$ for all $x\gt 0$, then in order to have $(2)$ for all $x\gt 0$, it is sufficient to have
$$2\rho\ge -2\lambda^2-2c,$$
i.e.
$$c\ge -\lambda^2-\rho$$If $f(x)=2\rho$ for some $x\gt 0$, then in order to have $(2)$ for all $x\gt 0$, it is sufficient to have
$$2\rho\gt -2\lambda^2-2c,$$
i.e.
$$c\gt -\lambda^2-\rho$$

*Case 2 : If $\alpha\gt -\frac 12$, then we have
$$\lim_{x\to 0^+}g(x)=+\infty$$
where $g(x)$ is the RHS of $(1)$. So, in order to have $(1)$ for all $x\gt 0$, at least we have to have
$$\lim_{x\to 0^+}f(x)=+\infty$$However, $\lim_{x\to 0^+}f(x)=+\infty$ does not necessarily imply that $(1)$ holds for all $x\gt 0$.

Conclusion :


*

*If $\alpha=-\frac 12$ and $f(x)\gt 2\rho$ for all $x\gt 0$, and $c\ge -\lambda^2-\rho$, then $H(x)\lt c$ holds for all $x\gt 0$.

*If $\alpha=-\frac 12$ and $f(x)\gt 2\rho$ for all $x\gt 0$, and $c\lt -\lambda^2-\rho$, then there exists an $x\gt 0$ such that $H(x)\ge c$.

*If $\alpha=-\frac 12$ and $f(x)=2\rho$ for some $x\gt 0$, and $c\gt -\lambda^2-\rho$, then $H(x)\lt c$ holds for all $x\gt 0$.

*If $\alpha=-\frac 12$ and $f(x)=2\rho$ for some $x\gt 0$, and $c\le -\lambda^2-\rho$, then there exists an $x\gt 0$ such that $H(x)\ge c$.

*If $\alpha\gt -\frac 12$ and $\displaystyle\lim_{x\to 0^+}f(x)=\gamma$ where $\gamma\in\mathbb R$, then there exists an $x\gt 0$ such that $H(x)\ge c$.

*If $\alpha\gt -\frac 12$ and $\displaystyle\lim_{x\to 0^+}f(x)=+\infty$, then whether $H(x)\lt c$ holds for all $x\gt 0$ depends on the form of $f(x)$. For example, if $f(x)=\frac{2\alpha+1}{x}+2\rho$ and $c\gt -\rho-\frac{\lambda^2}{2(\alpha+1)}$, then $H(x)\lt c$ holds for all $x\gt 0$. But if $f(x)=\frac{2\alpha+1}{\sqrt x}+2\rho$, then there does not exist $c\in\mathbb R$ such that $H(x)\lt c$ for all $x\gt 0$.)

Added : 
If $f(x) = \frac{2 \alpha + 1}{x} + \beta(x)$ where $\beta(0)=0$, then we see that $H(x)\lt c$ is equivalent to
$$\beta(x)\gt - \frac{\lambda^2}{\alpha+1}-2c$$
Whether $H(x)\lt c$ holds for all $x\gt 0$ depends on the form of $\beta(x)$.
For example, if $\beta(x)=2\rho-\frac{2\rho}{x+1}$ and $\rho\lt 0$, and $c\ge -\rho-\frac{\lambda^2}{2(\alpha+1)}$, then $H(x)\lt c$ holds for all $x\gt 0$.
But if $\beta(x)=\begin{cases}0&(x=0)\\2\rho-\frac 1x&(x\gt 0)\end{cases}$ and $\alpha\gt 0$, then there does not exist $c\in\mathbb R$ such that $H(x)\lt c$ for all $x\gt 0$. 
(I have not been able to find such a function $\beta(x)$ where $\beta(x)$ is continuous.)
