For any $\rho$ we want to minimize this function. The minimum of $\pi$ is obtained at a point where $x_2 = \frac{1}{2}x_{1}$ and where $x_1$ minimizes the function defined by $\begin{cases} \frac{1}{4}x_{1}^{2}-\rho(x_{1}-1),& \text{if} \ x_1 <1 \\\\ \frac{1}{4}x_{1}^2,&\text{if} \ x_{1} \geq 1 \end{cases}$
How do we know that for any $\rho >\frac{1}{2}$, $x_1 = 1$ is a minimum of the function? How did we get the $\frac{1}{2}$? By taking the derivative of those functions?
Added. We want the left branch to decrease if $x_1 <1$ and the right branch to increase for $x_1 \geq 1$. This means that $x_1=1$ is the minimum. It turns out for $\rho > \frac{1}{2}$ this always happens. How do we get the $\frac{1}{2}$?