For 1/A = 1/B + 1/C, why is A $\frac{1}{a}$  =$\frac{1}{b}$  +$\frac{1}{c}$  
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 $C$ $>1$

As you said, $\frac{1}{A}=\frac{C+B}{C*B}$
which means $A=\frac{B*C}{B+C}$
Now clearly $B > \frac{B*C}{B+C}$ because $\frac{B*C}{B+C} = B*\frac{C}{B+C}$ and $\frac{C}{B+C} < 1$.
Similarly, $C > \frac{B*C}{B+C}$ because $\frac{B*C}{B+C} = C*\frac{B}{B+C}$ and $\frac{B}{B+C} < 1$.
Thus $B > \frac{B*C}{B+C} = A$ and $C > \frac{B*C}{B+C} = A$
