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 $\theta > 0$.

$$
f(b,\theta)\geq 0
$$
\begin{align*}
 \implies\frac{2b-1}{b\sqrt{(1-\theta)(2b-1)^{2} + \theta}} + \frac{\sqrt{(1-\theta)(2b-1)^{2} + \theta} - 1}{2b^{2}(\theta - 1)}&\geq 0\\ \\
 \implies 2b-1+\frac{\left(b\sqrt{(1-\theta)(2b-1)^{2}+\theta}\right)\cdot\left(\sqrt{(1-\theta)(2b-1)^{2} + \theta} - 1\right)}{2b^{2}(\theta - 1)}&\geq 0\\\\
 \implies 2b-1+\frac{(1-\theta)(2b-1)^{2}+\theta-\sqrt{(1-\theta)(2b-1)^{2}+\theta}}{2b(\theta-1)}&\geq 0\\
 \implies 2b(\theta-1)(2b-1)+(1-\theta)(2b-1)^{2}+\theta-\sqrt{(1-\theta)(2b-1)^{2}+\theta}&\geq 0\\\\
\implies 2b\theta-2b+1-\sqrt{(1-\theta)(2b-1)^{2}+\theta}&\geq 0
\end{align*}
This implies the following :

*

*First Case : $b=0$ which is not the case since $b\in(0,1]$.


*Second Case : $0<b \leq 0.5$ and $\theta >\displaystyle \frac{4b^{2}-4 b+1 }{4  b^{2}-4  b}$


*Third Case : $0.5<b\leq 1$ and $\theta> 0$
Hence it follows that $\forall b\in(0,1]$ and $\forall\theta\in\mathbb{R}_{>0}$ with $\theta\neq 1$ we have that $f$ is non-negative.
