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 $C = 4$, this is the result we have so far ("enough" to produce a paper but it is probably not "optimal") but of course in any research project you hope to obtain the best possible result. 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.

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.

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 $C = 4$, this is the result we have so far ("enough" to produce a paper but it is probably not "optimal") but of course in any research project you hope to obtain the best possible result. 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 am so sorry that the initial post contains an important type-oh in the constants appearing in (1), which have been corrected for sure. If there is an satisfactory answer to this problem I will definitely put you (if you would like to tell us the real name of you) in the "Acknowledgement" part of the paper.

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 $C = 4$, this is the result we have so far ("enough" to produce a paper but it is probably not "optimal") but of course in any research project you hope to obtain the best possible result. 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.