How to show that $\frac{R_1R_2}{R_1+r_2}<(R_1,R_2)$ strictly using AM-GM inequality?

Question

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!!

Answer 1

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 $$ R_{eq} < \min_{k} R_k.$$

Answer 2

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 $R_{eq}<R_2$

Answer 3

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.