When resistances are connected in parallel the equivalent resistance is lower than the values of individual resistances in parallel?

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 $n\geq2$.