Solving inequality, but graphing/mathematica gives a completely different answer I have the following inequality problem, where I'm trying to solve for $\mu$. Here is the inequality:
$$\frac{1}{2} \left(-\sqrt{(c+d+\theta -\mu +1)^2+4 (c \theta -\theta  \mu )}+c+d+\theta -\mu +1\right) < 1$$
Here are all the constraints:
$$0<c<1, 0<d<1, 0<\theta<1$$
I thought this was relatively easy to answer, and got the following through basic algebra:
$$\mu < \frac{\theta + d}{1 + \theta} + c$$
However, when I try to graph this to check the results, I get a completely different answer. I tried it on Mathematica as well and got this, which seemed to work graphically as well:
$$\mu < c-2 \sqrt{d \theta +2 \theta ^2+\theta }+d+3 \theta +1$$
How did Mathematica get this result? I have no clue how the inequality in the beginning can be simplified to this.
Any help would be greatly appreciated!
 A: $$\frac{1}{2} \left(-\sqrt{(c+d+\theta -\mu +1)^2+4 (c \theta -\theta  \mu )}+c+d+\theta -\mu +1\right) < 1$$
Call $x = c+d+\theta+1$, then we have
$$\frac{1}{2} \left(-\sqrt{(x -\mu)^2+4 (c \theta -\theta  \mu )}+x-\mu\right) < 1$$.
Call now $x-\mu = y\rightarrow \mu = x-y$, to get
$$ \sqrt{y^2+4\theta y+\xi} > y-2$$,
where I defined $\xi = 4c\theta-4\theta x$. The solution is obtained by
$$
\begin{cases}y^2+4\theta y +\xi \geq 0\\ y-2<0\end{cases}\text{or}\begin{cases}y^2+4\theta y +\xi \geq 0\\ y-2\geq 0 \\ y^2+4\theta y +\xi>y^2+4-4y\end{cases}
$$
The $\Delta$ of the first quadratic inequality is
$$
16\theta^2-4\xi = 16\theta^2-16 c\theta+16\theta x=16\theta^2-16 c\theta+16\theta (c+d+1)+16\theta^2=16\theta (2\theta+d+1)
$$
which is $>0$ always. So we have
$$
\begin{cases}
y<2\\
y<y_2\text{ or }y>y_1
\end{cases}\text{or}
\begin{cases}
y\geq 2\\
y<y_2\text{ or }y>y_1\\
y>-\xi / (4\theta+4)
\end{cases}
$$
where $y_{1,2}=\frac{-4\theta\pm\sqrt{\Delta}}{2}$.
Now we have to mutually compare the values for $y$ ($2,y_1,y_2,-\xi/(4\theta+4)$), which requires solving other tedious inequalities. It turns out that the correct order is
$$
y_2<-\xi/(4\theta+4)<y_1<2
$$
for any choice of the parameters (where $y_2$ is the solution with the minus in front of the $\sqrt{\Delta}$). Therefore we have as a solution of the two systems $y<y_2$ or $y_1<y<2$ (left system) combined with $y\geq 2$ (right system). All together, $y<y_2$ or $y>y_1$. Since $y=x-\mu$, we have
$$
x-\mu<y_2\text{ or }x-\mu>y_1\rightarrow \mu> x-y_2\text{ or }\mu<x-y_1 .
$$
Simplifying for instance $x-y_1$ we get
$$
x-y_1=c+d+\theta+1+2\theta-\frac{1}{2}\times 4\sqrt{2\theta^2+\theta d+\theta}=c+d+1+3\theta-2\sqrt{2\theta^2+\theta d+\theta}
$$
(but please note that there is another branch of the solution, which you have omitted in your question...)
