My questions come from the proof of Theorem 5.14 in section 5.7 of this book. My first question can be stated as follows:
Suppose for positive numbers $V,y,\Delta,x,\delta >0$ we have $V \leq y^{-1}\left[2\Delta y^{-1} + \sqrt{2x} + \sqrt{2\delta}\right].$ Then if we define $$y^2 = 2K\epsilon^2\left[\left(\sqrt{x}+\sqrt{\delta}\right)^2 + \epsilon^{-1}K^{-\frac 12}\Delta + \sqrt{\delta\epsilon^{-1}K^{-\frac 12}\Delta} \right] \tag{1}$$ for $K >1$ and $\epsilon >0$, then authors claimed that the inequality for $V$ implies that $\epsilon V \leq K^{-\frac 12}$. To be honest, I have no clue as to the motivation of defining $y$ according to $(1)$. If I want to enforce the relation $V \leq K^{-\frac 12}/\epsilon$, I will just set up the relation $$y^{-1}\left[2\Delta y^{-1} + \sqrt{2x} + \sqrt{2\delta}\right] = K^{-\frac 12}/\epsilon,$$ which defines a quadratic equation in the variable $y^{-1}$ and whose solution can be explicitly found out. But this idea does not lead to the suggested expression for $y$...
My second question is again regarding details. Setting $\delta := \left((4\pi)^{-\frac 12} + \sqrt{z}\right)^2$ for fixed $z > 0$. The authors mentioned that by using repeatedly the elementary inequality $2ab \leq \theta a^2 + \theta^{-1}b^2$, we can derive the following upper bound from the identity/definition $(1)$ (recall that $K >1$ is fixed): $$K^{-\frac 12}y^2 \leq 2K\epsilon^2\left[\epsilon^{-1}\Delta + x + \sqrt{\epsilon^{-1}\Delta x} + \frac{2}{\sqrt{K} - 1}\left(\frac{1}{2\pi} + 2z\right)\right] \tag{2}$$ I tried a lot but still can not figure out the right way to arrive at $(2)$ from $(1)$, so any help is appreciated!