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The motivation of this question is that when you connect two or more resistors in parallel, then the total resistance is smaller than each of the resistors.

So in general, I guess I want to prove that given a set of numbers $\{a, b, c, \cdots, n\}$,$$\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \cdots + \frac{1}{n}\right)^{-1} \leq \min\{a, b, c, \cdots, n\}$$

I'm not sure if the inequality can be strict and I am guessing that the numbers in the set must also be positive.

For the case of two numbers, we have $$\left(\frac{1}{a} + \frac{1}{b}\right)^{-1} = \frac{ab}{a+b}$$ but it's not even obvious to me why this should be less than or equal to $a$ and $b$.

PhysicsMathsLove
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  • $n$ is the $14$th letter of the alphabet. Your question appears to be "If I have $14$ real numbers, add up their reciprocals, and then take the reciprocal, is that result smaller than the smallest of the original numbers?" If you want to ask something more general, it would make sense to label the numbers with subscripts, e.g. $a_1, a_2, \dotsc, a_n$. The way you have written the problem, it is kind of ambiguous. – Xander Henderson Sep 26 '20 at 21:16

1 Answers1

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$$a = \dfrac{1}{1/a}>\dfrac{1}{1/a+1/b+1/c+\ldots+1/n}=\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \cdots + \frac{1}{n}\right)^{-1}$$

Assuming $a,b,c,\ldots,n>0$

Daniel Huff
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  • The smallest nitpick in the whole world: if there is only one term then > is wrong and it should be $\geq$. – Randall Sep 26 '20 at 16:15