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?

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 $n\geq2$.
