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 $B = B(f)$?




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Remark: I think I have obtained a solution myself and maybe I can post my own solution later.



Obtaining the "best possible" inequality by tuning hyper-parameters
$\newcommand{\ga}{\gamma}$Letting $\ga\to\infty$ (with $A,B,\lambda,f(0)$ fixed), we see that the left-hand side of your inequality
\begin{equation}
\begin{aligned}
 &A + (\lambda + \gamma)^2 + \gamma^2 \\ 
 &\leq 2\,\sqrt{B + 2f(0)(\lambda - \gamma) + \lambda^2 + \gamma^2}\,\sqrt{A + (\lambda + \gamma)^2 + \gamma^2} \\
 &- \left(\gamma - f(0)\right)^2 
\end{aligned}
\tag{1} 
\end{equation}
is $\sim2\ga^2$, whereas the right-hand side of your inequality is $\sim(2\sqrt2-1)\ga^2<2\ga^2$.


So, there are no values of $A,B,\lambda,f(0)$ such that your inequality (1) holds for all real $\ga$.
Therefore, the inequality $A\le CB$ -- as well as, e.g., the inequality $A>CB$ -- hold for all real $C$ and all (actually nonexistent) $A,B$ such that your inequality (1) holds for some $\lambda,f(0)$ and all real $\ga$.


