Inequality for the combined resistance of two resistors connected in parallel

Solution 1:

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.