Simple And Elegant!

Let us assume that there is no friction. Then according to the energy conservation law, the velocity $v$ of the sliding body down the inclined plane from height $h$ at the foot is equal to the velocity which must be imparted to the body for its ascent to the same height $h$ . In this case, time of ascent is equal to time of descent.

If however, friction is taken into account, the velocity ${v}_{1}$ of the body at the end of descent is smaller than $v$ (due to work done against friction), while the velocity ${v}_{2}$ that has to be imparted to the body for raising it along the inclined plane is larger than $v$ for the same reason. Since the descent and ascent occur with constant(although different) accelerations, and the traversed path is same, the time ${t}_{1}$ of the descent and ${t}_{2}$ of ascent can be found from the formulas

$s=\frac{{v}_{1}{t}_{1}}{2} \text{ and }s=\frac{{v}_{2}{t}_{2}}{2}$ ,

where $s$ is the distance covered along the inclined plane. Since the inequality ${v}_{1}<{v}_{2}$ is satisfied, it follows that ${t}_{1}>{t}_{2}$ . Thus in the presence of sliding friction, time of descent is greater than time of ascent.