In the derivation of showing that a certain condition is sufficient and necessary, of which the context is given at the end, I couldn't rigorously resolve an intermediate step of technical nature.
Given three non-negative numbers $a$, $b$ and $c$ which satisfy $a\gg$ $\frac{1}{\frac{1}{b}+\frac{1}{c}}$, are we justified to put $a \gg b$ so that we have $$ \frac{1}{\frac{1}{b}+\frac{1}{a}} \approx b\quad ? $$
I'm tempted to just follow the line of \begin{align} \frac{1}{\frac{1}{b}+\frac{1}{a}} &= \frac{1}{\frac{1}{b}+ (\frac{1}{c} - \frac{1}{c}) + \frac{1}{a}} \\ &=\frac{1}{(\frac{1}{b}+ \frac{1}{c} + \frac{1}{a}) - \frac{1}{c}} \\ &\approx \frac{1}{(\frac{1}{b}+ \frac{1}{c}) - \frac{1}{c}} \\ &= b\ . \end{align}
I don't find this reasoning very convincing, since we don't seem to make a judgement on a possible differing order of magnitude between $b$ and $c$ and its resulting effect.
I've tried considering the following to shed some light.
Set $a = 10^{x}, b=10^y, c = 10^{z}$ and require $a \gg \frac{1}{\frac{1}{b} + \frac{1}{c}}$.
Taking the base 10 logarithm of both sides,
$$x \gg y+z-\log(b+c)\ .$$
Which doesn't clarify much.
Context. [All following quantities are, indeed, functions of frequency $\omega$. For readability, we omit explicitly denoting this dependence $f = f(\omega).]$
We consider the system of a coupled low-pass filter circuit.
Two low-pass filters are said to not load eachother if the input impedance $Z_{in,2}$ of the second one is much greater than the output impedance $Z_{out,1}$ of the first one.
If this is the case, we can approximate the system transfer function $H$ as the product of the respective transfer functions of the individual filters $H \approx H_1 \cdot H_2$.
It can be shown that the system transfer function is given by $$ H = \bigg( 1 - \frac{Z_{R_1}}{Z_{R_1} +\frac{1}{ \frac{1}{Z_{C_1}} + \frac{1}{Z_{R_2} + Z_{C_2}} } } \bigg) \cdot \frac{Z_{C_2}}{Z_{R_2} + Z_{C_2}}\ . $$ Similarly, we have $$ H_1 = \frac{Z_{C_1}}{Z_{C_1}+Z_{R_1}}\quad \text{ and }\quad H_2 = \frac{Z_{C_2}}{Z_{C_2} + Z_{R_2}} \ . $$
Now, $$Z_{in,2} \equiv Z_{C_2} + Z_{R_2} \gg \frac{1}{\frac{1}{Z_{R_1}} + \frac{1}{Z_{C_1}} } \equiv Z_{out,1}$$ seems to be a necessary and sufficient condition to justify $H \approx H_1 \cdot H_2$ .
In my opinion, the fact that $x \gg y+z-\log(b+c),$ where $a = 10^x,$ $b=10^y,$ and $c = 10^z,$ clarifies the problem almost completely.
We need only consider an example such as $y=100,$ $z=-100,$ where $\log(b+c) \approx \log(b) = y,$ to see that in this case $$y+z-\log(b+c) \approx z,$$ and therefore in this case the fact that $a\gg1/\left(\frac{1}{b}+\frac{1}{c}\right)$ has told us only that $a \gg c$ and has told us nothing about the relative sizes of $a$ and $b.$ We could have either $x = 0$ or $x = 200$ and all of the formulas above would still be true.
When I write that this "clarifies" the problem, by the way, I mean that this formula shows that the method of proof you are attempting does not work.