$\frac{1}{a}$ =$\frac{1}{b}$ +$\frac{1}{c}$
I read that $a$ is always < the smaller of $b$ and $c$
In the case of $0$ < $B$ $<1$ and $0$ < $C$ $<1$,
I can understand the rule as:
Firstly,
$$\frac{1}{A}=\frac{1}{B} +\frac{1}{C}=\frac{C+B}{B*C}$$
Now looking at 3 cases:
Case 1
$$\frac{1}{A}=\frac{1}{0.5} +\frac{1}{0.5}=\frac{0.5+0.5}{0.5*0.5}=\frac{1}{0.25}$$
Case 2
$$\frac{1}{A}=\frac{1}{0.2} +\frac{1}{0.9}=\frac{0.2+0.8}{0.2*0.8}=\frac{1}{0.04}$$
Case 3
$$\frac{1}{A}=\frac{1}{0.001} +\frac{1}{0.999}=\frac{0.001+0.999}{0.001*0.999}=\frac{1}{0.0009}$$
So if we look at the $B*C$ term, $$D=B*C$$
When $0$ < $B$ $<1$ and $0$ < $C$ $<1$
The smaller value (of $B$ or $C$) brings the product $B*C$ down to its level.
But how does the same hold true for when $B >1$ and $C>1$
Case 4
$$\frac{1}{A}=\frac{1}{1} +\frac{1}{1}=\frac{1+1}{1*1}=\frac{2}{1}$$
Hence $$A=1/2$$
Case 5
$$\frac{1}{A}=\frac{1}{5} +\frac{1}{5}=\frac{5+5}{5*5}=\frac{10}{25}$$
Hence $$A=25/10=2.5$$
I can see in both of these cases $A$ is still $<B$ and $A<C$. I know there is something going on here in relation to the A value and the result on the RHS of the equation but I can't quite put my finger on what causes $A$ to be less than $B$ and less than $C$ where $B$ and $C$ $>1$
Since everything is strictly positive, it's simply the fact that $$\frac1A=\frac1B+\frac1C>\frac1B$$ And $$\frac1A>\frac1B\implies B>A$$ Same for $C$. Since $A$ is smaller than boh, it is smaller than the minimum.
I don't really understand why in you decided to consider only three numbers.