The question is from "Kiselev's Geometry: Planimetry". You can find the question here: http://schoolnova.org/classes/f2010/math7/assignments/Kiselev-pg-$41-48$.pdf. Page $41$, problem $94$ of the book. I just copied the question word for word.
Tried to do the problem first with an equilateral triangle of side with 1 unit and the centroid as a point. But the example disproves the statement. Semiperimeter $(1+1+1)\times0.5=1.5$ , and the sum of segments $(\frac{\sqrt{3}}{2}\times3\times{\frac{2}{3}})=1.73.$ Therefore sum of segments is GREATER than semiperimeter.
Am I wrong with my example or is the book wrong on this problem?
Yes, the book appears to be incorrect!
Here's a proof of the inequality in the opposite direction, for any triangle and any point (whether in the triangle or not).
Let $A$, $B$, $C$, be the vertices of the triangle, with corresponding perimeter segments $AB, BC, AC$. Let $\bar{A}, \bar{B}, \bar{C}$ be the segments connecting the vertices to any point in the plane (doesn't need to be inside the triangle!). The triangle inequality gives that $\bar{A} + \bar{B} \leq AB$, $\bar{B} + \bar{C} \leq BC$, $\bar{A} + \bar{C} \leq AC$. Summing these inequalities, and noting that strict inequality must apply to at least one of them if the triangle has nontrivial angles, you've proved precisely that the sum of the segments connecting the point to the vertices is greater than the semiperimeter.
Now, see if you can prove that the sum of those segments is less than the perimeter. This is also a true statement!