Why is equivalent resistance in parallel circuit always less than each individual resistor? There are $n$ resistors connected in a parallel combination given below.

$$\frac{1}{R_{ev}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}+\frac{1}{R_{4}}+\frac{1}{R_{5}}.......\frac{1}{R_{n}}$$
Foundation Science - Physics (class 10) by H.C. Verma states (Pg. 68)

For two resistances $R_{1}$ and $R_{2}$ connected in parallel,
$$\frac{1}{R_{ev}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}=\frac{R_{1}+R_{2}}{R_{1}R_{2}}$$ $$R_{ev}=\frac{R_{1}R_{2}}{R_{1}+R_{2}}$$
We see that the equivalent resistance in a parallel combination is less than each of the resistances.

I observe this every time I do an experiment on parallel resistors or solve a parallel combination problem.
How can we prove $R_{ev}<R_{1},R_{2},R_{3},...R_{n}$ or that $R_{ev}$ is less than the Resistor $R_{min}$, which has the least resistance of all the individual resistors?
 A: Think about current flow.
If we take each individual resistor and determine the current for the applied voltage, we get: $$I_T=\frac {V}{R_1} +\frac {V}{R_2} + ...$$
Dividing everything by the voltage give us:
$$\frac {I_T}{V}=\frac {1}{R_1} +\frac {1}{R_2} + ...$$
Which is the same as:
$$\frac {1}{R_{eq}}=\frac {1}{R_1} +\frac {1}{R_2} + ...$$
Since there is more current flowing in all the resistors than through just one resistor, then the equivalent resistance must be less than the individual resistors.
A: The individual resistances are all positive, so the sum
$$ \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots \,,$$
is larger than the inverse of any of the individual resistances, and that means that the inverse of the sum is necessarily smaller than any of the resistances.
No mucking around with the two-resistor form required.
A: We can prove it by induction. Let 
$$
\frac{1}{R^{(n)}_{eq}} = \frac{1}{R_1} + \cdots+ \frac{1}{R_n}
$$
Now, when $n=2$, we find
$$
\frac{1}{R^{(2)}_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \implies R_{eq}^{(2)} = \frac{R_1 R_2}{R_1+R_2} = \frac{R_1}{1+\frac{R_1}{R_2}} = \frac{R_2}{1+\frac{R_2}{R_1}}
$$
Since $\frac{R_1}{R_2} > 0$, we see that $R^{(2)}_{eq} < R_1$ and $R^{(2)}_{eq} < R_2$ or equivalently $R^{(2)}_{eq} < \min(R_1, R_2)$. 
Now, suppose it is true that $R^{(n)}_{eq} < \min (R_1, \cdots, R_n)$. Then, consider
$$
\frac{1}{R^{(n+1)}_{eq}} = \frac{1}{R_1} + \cdots+ \frac{1}{R_n} + \frac{1}{R_{n+1}} = \frac{1}{R^{(n)}_{eq}} + \frac{1}{R_{n+1}}
$$
Using the result from $n=2$, we find
$$
R^{(n+1)}_{eq} < \min ( R_{n+1} , R^{(n)}_{eq} ) < \min ( R_{n+1} , \min (R_1, \cdots, R_n))
$$
But
$$
\min ( R_{n+1} , \min (R_1, \cdots, R_n)) = \min ( R_{n+1} ,  R_1, \cdots, R_n)
$$
Therefore
$$
R^{(n+1)}_{eq} < \min ( R_1, \cdots, R_n , R_{n+1} )
$$
Thus, we have shown that the above relation holds for $n=2$, and further that whenever it holds for $n$, it also holds for $n+1$. Thus, by induction, it is true for all $n\geq2$.
A: It might be easier to think in terms of conductance which is the inverse of resistance.
The more paths there are between A and B which conduct electricity, the greater is the amount of current which can flow - ie the greater is the conductance of the network. The total conductance is greater than that of any individual path, because each additional path always increases the amount of electricity which can be conducted, it never reduces it. In particular, the total conductance is always greater than the largest individual conductance.
Translating this back in terms of resistance R (which is the inverse of conductance S - ie R = 1/S), the total resistance is smaller than the smallest individual resistance.  
