How do we prove the proposed expression is nonnegative? So the expression which I am interested in is given by
\begin{align*}
f(b,\theta) = \frac{2b-1}{b\sqrt{(1-\theta)(2b-1)^{2} + \theta}} + \frac{\sqrt{(1-\theta)(2b-1)^{2} + \theta} - 1}{2b^{2}(\theta - 1)}
\end{align*}
where $b\in(0,1]$, $\theta\in\mathbb{R}_{>0}$ and $\theta\neq 1$.
It is part of my research project and I would like to know a good way to approach it.
Any hint is appreciated.
EDIT
According to WA, the solution is given by $0 < b \leq 1$ when $\theta > 0$.

Denote $A = (1-\theta)(2b-1)^2 + \theta$.
Since $(2b-1)^2 \le 1$, we have $A \ge (1-\theta)(2b-1)^2 + \theta (2b-1)^2 = (2b-1)^2$.
We have
\begin{align}
f(b, \theta) &= \frac{2b-1}{b\sqrt{A}} + \frac{\sqrt{A} - 1}{2b^2(\theta - 1)} \\
&= \frac{2b-1}{b\sqrt{A}} + \frac{A - 1}{2b^2(\theta - 1)(\sqrt{A} + 1)}\\
&= \frac{2b-1}{b\sqrt{A}} + \frac{2(1-b)}{b(\sqrt{A} + 1)}\\
&= \frac{2b - 1 + \sqrt{A}}{b\sqrt{A}(\sqrt{A} + 1)}\\
&\ge \frac{2b - 1 + \sqrt{(2b-1)^2}}{b\sqrt{A}(\sqrt{A} + 1)}\\
&\ge 0.
\end{align}
We are done.
