Deriving a short inequality using the indicator function I am struggling with an inequality in this paper, which is stated right after equation (5.10).
For $0<\delta<1$ and a constant $K_1>0$, there exist positive constants $K_2$ and $b$ such that
$$(1-\delta)\left[a+ 2\sqrt{a} \cdot \sqrt{K_1} + K_1\right]\leq (1-\delta/2)\cdot a + b \cdot \mathbb{I}_{\{a \leq K_2\}}, $$
where $\mathbb{I}$ denotes the indicator function.
Intuitively it makes sense, that we increase the right side enough as soon as $a$ is large enough by taking $\delta/2$. However, I do not see how we can determine the constants $b$ and $K_2$. 
I am thankful for any hints.
 A: When we reformulate the left hand side, we get
\begin{align*}
&(1-\delta) \cdot [a + 2 \sqrt{a} \cdot \sqrt{K_2} + K_2] = (1-\delta/2 - \delta/2) \cdot [a + 2 \sqrt{a} \cdot \sqrt{K_2} + K_2] \\
& = (1-\delta/2)a + (1-\delta/2)(2 \sqrt{a} \sqrt{K_2} + K_2) - \frac{\delta}{2} (a + 2 \sqrt{a} \sqrt{K_2} + K_2)\\
& = (1-\delta/2)a + (1- \delta) (2 \sqrt{a} \sqrt{K_2} + K_2) - \frac{\delta}{2}a =: (\star)
\end{align*}
We want to bound everything after the first term, which we will denote in the following by a function $g: \mathbb{R}^3 \rightarrow \mathbb{R}, (\delta, a, K_2) \mapsto (1- \delta) (2 \sqrt{a} \sqrt{K_2} + K_2) - \frac{\delta}{2}a$. In the case that $g \leq 0$, we can bound $(\star)=(1-\delta/2)a + g(\delta, a, K_2) \leq (1-\delta/2)a$. We can determine a constant 
\begin{align*}
K_3:= K_2 \cdot 2(\delta-1) \cdot \left[ 2 \sqrt{2} \sqrt{\frac{\delta-2}{\delta-1}} (1-\delta) + 3\delta - 4 \right]^{-1}
\end{align*}
such that $g(\cdot, a, \cdot) \leq 0$ if $a \geq K_3$ (Wolframalpha helped me here). Looking at the set $\Omega := [0,1]\times[0, K_3]$, we note that the function $g_{|\Omega}$ is continuous on a compact set ($K_2$ is not a problem, as it really is a constant) and therefore attains a maximum. If we define this maximum as $b$, we get  the desired result.
