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I was reading about parallel circuits in Physics.Equivalent resistance of $n$ resistors in parallel is given by $\displaystyle\frac1{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+...+\frac{1}{R_n}$.

I tried to prove that $R_{eq}$ will always be less than $R_1,R_2,...,R_n$.

I tried to prove it for two resistors,where $R_{eq}=\frac{R_1R_2}{R_1+R_2}$.

By applying AM-GM on $R_1,R_2$we have,

$\frac{R_1R_2}{4}\geq \frac{R_1R_2}{R_1+R_2}$.

Now,I have no idea how to show from here that $(R_1,R_2)\geq\frac{R_1R_2}{R_1+R_2}$ and how it can be extended for $R_n$

Thanks for any help!!

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  • $\begingroup$ FYI: The harmonic mean of a set of numbers $\{a_n\}$ is $H=\left(\sum_n \frac{1}{a_n}\right)^{-1}$. $\endgroup$ – Semiclassical Jun 3 '16 at 17:12
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    $\begingroup$ $=n\sum_n\Bigl(\frac1{a_n}\Bigr)^{-1}$ more exactly. $\endgroup$ – Bernard Jun 3 '16 at 17:15
  • $\begingroup$ This is not mathematical physics. Edited tags. $\endgroup$ – user_of_math Jun 3 '16 at 17:27
  • $\begingroup$ $\displaystyle{{1 \over R_{\mathrm{eq}}} > {1 \over R_{k}}\,,\quad\forall\ k}$. $\endgroup$ – Felix Marin Jun 4 '16 at 5:22
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That is more trivial. If $$ \frac{1}{R_{eq}}=\sum_{k=1}^{n}\frac{1}{R_k} $$ obviously $$ \frac{1}{R_{eq}}> \frac{1}{R_k} $$ for any $k\in\{1,2,\ldots,n\}$, hence $R_{eq}< R_k$, so $$ R_{eq} < \min_{k} R_k.$$

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    $\begingroup$ @tatan: if $ u = a+b+c+\ldots $ and $a,b,c,\ldots >0$, then $u>a, u>b, u>c,\ldots$. If that isn't trivial... $\endgroup$ – Jack D'Aurizio Jun 3 '16 at 17:39
  • $\begingroup$ Thanks for your answer....but I was looking for AM-GM.... $\endgroup$ – tatan Jun 3 '16 at 17:47
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suppose $1/R_1 = x$ and $1/R_2=y$ and $1/R_{eq}=z$ now we know that $z=x+y$ and $x>0,y>0$

therefore $z>x$ and $z>y$ and hence $1/R_{eq}>1/R_1 $ and $1/R_{eq}>1/R_2 $

and therefore $R_{eq}<R_1$ and $R_{eq}<R_2$

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Suppose we remove one of the resistors $R_k$ from the circuit. Then the new equivalent resistance will $\left(\frac{1}{R_{eq}}-\frac{1}{R_k}\right)^{-1}>R_{eq}$ i.e. the equivalent resistance strictly increases. If we repeat this until only one resistor $R_f$ remains, then $R_f$ is the final equivalent resistance. Hence each individual resistor must have a larger resistance than the equivalent resistance of them in parallel.

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