Answer
The equivalent resistance $R_{12}$ of $R_1$ and $R_2$ is less than $R_1$
Work Step by Step
We can find the equivalent resistance when the resistors are in parallel:
$\frac{1}{R_{12}} = \frac{1}{R_1}+\frac{1}{R_2}$
$\frac{1}{R_{12}} = \frac{R_2}{R_1~R_2}+\frac{R_1}{R_1~R_2}$
$\frac{1}{R_{12}} = \frac{R_1+R_2}{R_1~R_2}$
$R_{12} = \frac{R_1~R_2}{R_1+R_2} \lt \frac{R_1~(R_1+R_2)}{R_1+R_2} = R_1$
The equivalent resistance $R_{12}$ of $R_1$ and $R_2$ is less than $R_1$.
