Resist the Resistances

Let two resistances be A and B .

So, S=A+B and P= $\frac { AB }{ A+B }$ .

$\because \quad S=nP\\ A+B=\frac { nAB }{ A+B }$

${ (A+B) }^{ 2 }=nAB\\ { A }^{ 2 }+{ B }^{ 2 }+2AB=nAB$

now dividing both sides with AB, we get ,

$\therefore \frac { { A }^{ 2 } }{ AB } +\frac { { B }^{ 2 } }{ AB } +2=n$

$\frac { A }{ B } +\frac { B }{ A } =n-2$

Here, using $A.M.\ge G.M.$ on $\quad \frac { A }{ B } \quad and\quad \frac { B }{ A }$ we get,

$\frac { \frac { A }{ B } +\frac { B }{ A } }{ 2 } \ge \sqrt { \frac { A }{ B } \times \frac { B }{ A } }$

So, $\frac { A }{ B } +\frac { B }{ A } \ge 2\\ n-2\ge 2\\ n\ge 4$