# The largest and smallest value with 4 resistors

In physics, a resistor is an electrical component that inhibits current. We represent its ability to inhibit current with a value called its resistance, represented with $R$. ($R$ is always positive.)

There are two ways to combine two resistors to make another resistor: in parallel, or in series. If combined in series, $R_{\mathrm{new}} = R_1 + R_2$. If combined in parallel, $\frac1{R_{\mathrm{new}}} = \frac1{R_1} + \frac1{R_2}$.

Now the actual puzzle:

Given 4 different resistors, let $L$ be the largest resistance you can make with those resistors and let $S$ be the smallest resistance you can make with those resistors. Show that $L \geq 16S$.

• I've edited your question to add an explanation of the "rules", since not everyone knows resistance formulas. – Deusovi Jul 22 '16 at 12:57
• Isn't that rather a math's question than a puzzle? Seems straight forward calculation is all that is needed... – BmyGuest Jul 22 '16 at 14:48

If you put all four resistors in series, you get a resistance $R_s=R_1+R_2+R_3+R_4$. If you put all four in parallel, you get a resistance $R_p$ such that $\frac1{R_p}=\frac1{R_1}+\frac1{R_2}+\frac1{R_3}+\frac1{R_4}$.

Now $\frac{L}S\ge\frac{R_s}{R_p}=\left(R_1+R_2+R_3+R_4\right)\left(\frac1{R_1}+\frac1{R_2}+\frac1{R_3}+\frac1{R_4}\right)$.

By AM-GM, $$\left(R_1+R_2+R_3+R_4\right)\left(\frac1{R_1}+\frac1{R_2}+\frac1{R_3}+\frac1{R_4}\right)\\\ge\left(4\sqrt[4]{R_1R_2R_3R_4}\right)\left(4\frac1{\sqrt[4]{R_1R_2R_3R_4}}\right)=16$$ as desired. Equality holds iff $R_1=R_2=R_3=R_4$.

First consider the case of two resistors.

By the Arithmetic-Geometric Mean Inequality Theorem, given two positive reals $a, b$, $\sqrt{ab}\leqslant \frac{a+b}2$. Applying that theorem both to the resistances and to their reciprocals,

$\sqrt{ab}\leqslant \dfrac{a+b}2=\dfrac{R_s}2$

$\dfrac1{\sqrt{ab}}\leqslant \dfrac{1/a+1/b}2$

$\Rightarrow \sqrt{ab}\geqslant \dfrac2{1/a+1/b}=2R_p(a,b)$

$\Rightarrow 4R_p(a,b)\leqslant R_s$

where $R_p(a,\dots)$ denotes the resistance of resistors $a,\dots$ in parallel. Moving on now to four resistors, we have

$16R_p(a,b,c,d)\leqslant 4(R_p(a,b)+R_p(c,d))\leqslant a+b+c+d$

Proof of the Arithmetic-Geometric Mean Inequality Theorem for two reals $a, b$. If $a\leqslant b$, then

$a(b-a)\leqslant b(b-a)$

$\Rightarrow ab-a^2\leqslant b^2-ab$

$\Rightarrow 2ab\leqslant a^2+b^2$

$\Rightarrow 4ab\leqslant a^2+2ab+b^2=(a+b)^2$

$\Rightarrow ab\leqslant (\frac{a+b}2)^2$

$\Rightarrow \sqrt{ab}\leqslant \frac{a+b}2$

If $b\leqslant a$ the same result follows. Therefore it is true for all real $a, b$.

Actually there is no need to appeal to the AM-GM inequality.

Given that $L=R_1+R_2+R_3+R_4$

and that $\frac{1}{S}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}$,
we obtain
$$\frac{L}{S}=(R_1+R_2+R_3+R_4)\left(\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}\right)=4+\frac{R_2+R_3+R_4}{R_1}+\frac{R_1+R_3+R_4}{R_2}+\frac{R_1+R_2+R_4}{R_3}+\frac{R_1+R_2+R_3}{R_4}$$ Then we use the function inequality valid for $x>0$ $$f(x)=x+\frac{1}{x}= 2+(\sqrt{x}-\frac{1}{\sqrt{x}})^2\geq 2$$ Apply this then to $x = \frac{R_1}{R_2}, \frac{R_1}{R_3},\dots$ in the expression above and one obtains $$\frac{L}{S}\geq 16$$

Sorry, but the inequality seems wrong...

The allowed uses of the resistors are not made explicit in the -original-question.

If someone gives me four resistors, there is a resistance value I can make -with those resistors- that the other answers do not account for. By using only the resistor legs I can make a very low resistance.

This low resistance is independent of the resistor-values. So any inequality describing the situation should include the specific resistor values. Therefore it must be wrong.

Perhaps a trick question to separate engineers from theoreticians?