In the comments you asked me by name for a calculus way. Personally I would not do it this way.
Let us assume that the stick is broken at two random points $X, Y$ where the two points are independently uniformly distributed on $[0,l]$, so pieces with lengths $\min(X,Y), \max(X,Y)-\min(X,Y), l-\max(X,Y)$ and a triangle cannot be constructed if any of the lengths is $l/2$ or more (almost certainly mutually exclusive events).
Then $\displaystyle \Pr(\min(X,Y) \ge l/2)=\int_{x=l/2}^{l} \int_{y=l/2}^{l} \frac1{l^2} \,dy \,dx = \frac14$
$\displaystyle \Pr\left(\max(X,Y)-\min(X,Y) \ge l/2\right)=\int_{x=l/2}^{l} \int_{y=0}^{x-l/2} \frac1{l^2} \,dy \,dx + \int_{x=0}^{l/2} \int_{y=x+l/2}^{l} \frac1{l^2} \,dy \,dx = \frac14$
$\displaystyle \Pr(l-\max(X,Y) \ge l/2)=\int_{x=0}^{l/2} \int_{y=0}^{l/2} \frac1{l^2} \,dy \,dx = \frac14$.
The probability you want is $1-\dfrac14-\dfrac14-\dfrac14=\dfrac14$.