Inequality for the combined resistance of two resistors connected in parallel Exact question: In electric circuit theory, the combined resistance $R$ of two resistors $R_1>0$ and $R_2>0$ connected in parallel obeys
$$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}$$
Show that $R<\sqrt{\frac{R_1R_2}{2}}$.  This proves that $R$ cannot be large if the individual resistances $R_1$ and $R_2$ are small.
 A: Using AM,GM inequality we have $$\frac{\frac1{R_1}+\frac1{R_2}}2\ge\sqrt{\frac1{R_1R_2}}$$
A: We have 
$$R=\frac{R_1R_2}{R_1+R_2}.\tag{1}$$
Note that 
$$R_1+R_2=(\sqrt{R_1}-\sqrt{R_2})^2+2\sqrt{R_1R_2}.\tag{2}$$
The right-hand side of (2) is clearly $\ge 2\sqrt{R_1R_2}$. It follows from (1) that
$$R\le \frac{R_1R_2}{2\sqrt{R_1R_2}}=\frac{\sqrt{R_1R_2}}{2}.$$ 
Remark: This is a stronger inequality than the one in the post.
Another way:  Since you mentioned your Calc I class, let us use 
calculus. Let the product $R_1R_2$ be fixed, say equal to $k$. We then have
$$R=\frac{k}{R_1+R_2}.$$ 
To use more familiar notation, write $x$ instead of $R_1$, and $y$ instead of $R_2$. We want to calculate the maximum of
$$R(x)=\frac{k}{x+y},$$
given $xy=k$, in terms of $k$.  So we want to minimize $x+y$. That means we want to minimize $f(x)$, where
$$f(x)=\frac{1}{x}+\frac{x}{k},$$
and $x$ ranges over the positive reals.
Note that 
$$f'(x)=-\frac{1}{x^2}+k.$$
We can now  see that the minimum is reached at $x=\frac{1}{\sqrt{k}}$, and that the minimum value is $\frac{2}{\sqrt{k}}$. It follows that the maximum value of $R$ given $xy=k$ is equal to $\frac{\sqrt{k}}{2}$. That is precisely the result that was proved above without calculus.      
