Rigorous proof that $1/\sum_ir_i^{-1}\le\min r_i$

Question

I am taking an introductory physics course where the equivalent resistance of resistors in parallel is: $$\bigg ( \sum_{i=0}^n \frac 1{r_i} \bigg ) ^{-1} = R_{eq}$$ My book says that $R_{eq}$ will always be less than or equal to the least of the other $r's$, (assuming all resistances are positive and nonzero, of course) and I can reason out how this would work, but I have not seen a rigorous proof of this fact and I think seeing one would be helpful. Thanks in advance.

Answer 1

HINT: Assume $n \ge 2$ first, as this is not interesting for $n=1$. Then:

$$\Big(\sum_{i=1}^n \frac{1}{r_i}\Big) > \max \Big\{\frac{1}{r_1},\frac{1}{r_2}, \ldots, \frac{1}{r_n} \Big\} = \frac{1}{\min\{r_1,\ldots, r_n\}}.$$ [Indeed, the first "$>$" from noting that the sum of $n$ positive numbers is greater than the max of these $n$ numbers, and the "$=$" from noting that $\frac{1}{x}$ is a decreasing function in $x$ where $x$ is a positive real number.] Thus, raising both sides of the above to the $(-1)$-th power [and using the observations that $a>b$; $a,b >0$; $\implies a^{-1}<b^{-1}$, and $\Big(\frac{1}{y}\Big)^{-1} = y$] yields:

$$ \Big(\sum_{i=1}^n \frac{1}{r_i}\Big)^{-1} < \Big(\frac{1}{\min\{r_1,\ldots, r_n\}}\Big)^{-1} = \min\{r_1,\ldots, r_n\}.$$

Answer 2

If the minimum is $r_m$, your inequality $$ \left(\sum_{i=1}^n \frac{1}{r_i} \right)^{-1} \le r_m $$ is equivalent to $$ 1 \le r_m \left(\sum_{i=1}^n \frac{1}{r_i} \right) $$

which is certainly true since one of the terms on the right is $r_m/r_m = 1$ and the others are positive. Moreover the inequality is strict if $n > 1$.