Well, I've the following system of equations:
$$ \begin{cases} \text{f}=\frac{\text{R}_3+\text{R}_4}{\text{R}_3}\\ \\ \text{d}=\pi\cdot\text{h}\cdot\left\{\text{C}_1\cdot\left(\text{R}_1+\text{R}_2\right)-\frac{\text{C}_2\cdot\text{R}_1\cdot\text{R}_4}{\text{R}_3}\right\}\\ \\ \text{h}=\frac{1}{2\pi\cdot\sqrt{\text{C}_1\cdot\text{C}_2\cdot\text{R}_1\cdot\text{R}_2}} \end{cases}\tag1 $$
Context of the question: I'm designing a analog electronic filter and this are the parameters of the filter. So all the variables has to be real and positive.
Question: Someone wrote to me that out of the system of equations follows the following condition:
$$\text{f}-1<\frac{\text{C}_2}{\text{C}_1}\le\text{d}^2+\text{f}-1\tag2$$
And I do not see how that follows?!
My work:
In order to have all the variables real and positive we know from the second equation in the system of equations that we need the condition:
$$\text{C}_1\cdot\left(\text{R}_1+\text{R}_2\right)>\frac{\text{C}_2\cdot\text{R}_1\cdot\text{R}_4}{\text{R}_3}\tag3$$
But that does not help (I think) to find that the condition in $\left(2\right)$ is right :(.
The above statement (2) is not true. $f-1=R_4/R_3$. We can rewrite $d$ as $$d=\pi h C_1R_1\left[(1+\frac{R_2}{R_1})-(f-1)\frac{C_2}{C_1}\right]$$ Let's take $f-1=2$, and $C_2/C_1=1$. If $R_2>R_1$, then I have some positive $d$ and $h$ values such as that the first inequality in (2) is not true.