Let $EFG$ be a triangle such that $P$, $N$ and $M$ are the midpoints of $EF$, $EG$ and $FG$ respectively. The problem is to show that:
$$Q/2< EM+FN+GP<Q,$$ where $Q$ is the perimeter of the triangle $EFG$.
My approach: I figured that I'd try using the triangle inequality in the triangles formed inside, but that didn't lead me to anything.
So any hint or help are much appreciated.
Properly D S suggests: \begin{align*} 2EM<EG+EF,\\ 2GP<EG+GF,\\ 2FN<EF+FG. \end{align*} By addition, $2(EM+FN+GP)<2(EF+FG+EG)\ \Rightarrow\ EM+FN+GP<Q$.
Let $O$ be the centroid of $\triangle EFG$. The left inequality can be obtained like this: \begin{align*} OE+OF>EF\Rightarrow\ 2EM/3+2FN/3>EF,\\ OE+OG>EG\Rightarrow\ 2EM/3+2GP/3>EG,\\ OF+OG>EG\Rightarrow\ 2FN/3+2GP/3>FG.\\ \end{align*} By addition, $$ 4(EM+FN+GP)/3>EF+FG+EG\ \Rightarrow\ EM+FN+GP>3Q/4>Q/2. $$ This last inequality is in the book on page 86. However, the first inequality is not there.
This last inequality can also be proved like this: \begin{align*} OE+ON>EN\Rightarrow\ 2EM/3+FN/3>EG/2,\\ OF+OP>FP\Rightarrow\ 2FN/3+GP/3>EF/2,\\ OG+OM>GM\Rightarrow\ 2GP/3+EM/3>FG/2.\\ \end{align*} By addition, $$ EM+FN+GP>(EF+FG+EG)/2\ \Rightarrow\ EM+FN+GP>Q/2. $$