I encountered the following problem in one of my research projects which can be encapsulated as follows. Let's say we have a set $\mathcal{C}$ of functions $f$ defined from $\mathbb R_+$ to $\mathbb R$, and we have two functionals $A = A(f) \,\colon\, \mathcal{C} \to \mathbb{R}_+$ and $B = B(f) \,\colon\, \mathcal{C} \to \mathbb{R}_+$. Assume that we have the following inequality $$ A + (\lambda + \gamma)^2 + \lambda^2 \leq 2\,\sqrt{B + 2f(0)(\lambda - \gamma) + \lambda^2 + \gamma^2}\,\sqrt{A + (\lambda + \gamma)^2 + \lambda^2} - \left(\gamma - f(0)\right)^2 \tag{1},$$ which holds for every $\lambda,\gamma \in \mathbb R$. Additional, we also know that $f^2(0) \leq \min\{\frac{3}{4}A, \frac{1}{3}B \}$ (so $f(0)$ is not really "free"). The problem is to find the smallest fixed (or universal) constant $C > 0$ for which $$ A \leq C\,B \tag{2}$$ holds independently of the choice of $f \in \mathcal{C}$. For instance, one baby special case is when $\lambda = \gamma = 0$, and we can easily see that $(2)$ holds with $C = 4$. So my question boils down to what is the best possible constant in (2) that one can deduce from (1) by tuning $\lambda$ and $\gamma$ while keeping in mind that we have upper bounds on $f^2(0)$ in terms of $A = A(f)$ and $B = B(f)$?
Remark: I think I have obtained a solution myself and maybe I can post my own solution later.