Not only that the perimeter of $ DEF$ is less than $ s/2$, but $ DEF$ is the inscribed triangle with the smallest perimeter. This result is known as Fagnano's theorem/problem. Several approaches are possible. For example, see: http://forumgeom.fau.edu/FG2007volume7/FG200728.pdf or http://forumgeom.fau.edu/FG2004volume4/FG200422.pdf. However, the most beautiful proof for this is an older one due to Schwarz, by reflecting the triangle $ ABC$ around the angles $ A$ and $ B$ (respectively). In this case, we see that the image $ B'C'$ of $ BC$ after these 6 rotations is parallel to $ BC$ and the line $ D - F - .... - D'$ (where $ D'$ is the image of $ D$) is nothing else but twice the perimeter of $ DEF$. Now since twice the perimeter of any other inscribed triangle will be a polygonal line (and not straight line as in the case of $ DEF$), it will be less than the $ DD'$. This proves our problem.

http://www.artofproblemsolving.com/viewtopic.php?t=245840

We need to prove that $a*cosA+b*cosB+c*cosC\leq \frac{1}{2}(a+b+c)$, where $a,b,c$ are the sides of $ABC$. It is equivalent to $\frac{ab^2+ac^2-a^3}{bc}+\frac{bc^2+ba^2-b^3}{ca}+\frac{ca^2+cb^2-c^3}{ab}\leq \frac{1}{2}(a+b+c)$. We can rewrite this as $2(a^2b^2+b^2c^2+c^2a^2) \leq (a+b+c)abc + a^4+b^4+c^4$, and then as $(a+b+c)^4+9abc(a+b+c) \geq 4(a+b+c)^2(ab+bc+ca)$, which is Schur's inequality.

$a\cdot cos(\alpha)+b\cdot cos(\beta)+c\cdot cos(\gamma)\leq \dfrac{1}{3}(cos(\alpha)+cos(\beta)+cos(\gamma))(a+b+c)\leq \dfrac{a+b+c}{2}=s$

My solution: Let $ H $ be the orthocenter of $ \triangle ABC $ . Let $ M $ be the midpoint of $ BC $ and $ E' $ be the reflection of $ E $ in $ BC $ . Since $ H $ is the incenter of $ \triangle DEF $ , so $ BC $ is the external bisector of $ \angle EDF $ , hence we get $ E' \in DF $ . Since $ MF=ME=\frac{BC}{2}=ME' $ , so we get $ DE+DF=DE'+DF=E'F \leq MF+ME'=BC $ . Similarly, we can prove $ EF+ED \leq CA $ and $ FD+FE \leq AB $ , so combine three inequalities we get $ EF+FD+DE \leq \frac{AB+BC+CA}{2} $ . Q.E.D

WLOG $a\geq b\geq c\implies \alpha\geq \beta \geq \gamma \implies \cos\alpha \leq \cos\beta \leq \cos\gamma$. By rearrangement iequality: $a \cdot\cos\alpha+b \cdot\cos\beta+c\cdot\cos\gamma\leq b \cdot\cos\alpha+c\cdot \cos\beta+a\cdot\cos\gamma$. And $a \cdot\cos\alpha+b \cdot\cos\beta+c\cdot\cos\gamma\leq c \cdot\cos\alpha+a \cdot\cos\beta+b\cdot\cos\gamma$. $\implies a\cdot\cos\alpha+b \cdot\cos\beta+c\cdot\cos\gamma\leq \frac{b\cdot\cos\gamma +c\cdot \cos\beta}{2}+\frac {a\cdot\cos\gamma + c\cdot\cos\alpha}{2}+\frac{a\cdot\cos\beta+b \cdot\cos\alpha}{2}=\frac{a+b+c}{2}$.

Let $O$ be the circumcenter of triangle $ABC$. $R=AO$. Then easily we'll have $AO \perp EF,BO \perp DF,CO \perp DE$. So $2S_{ABC}=R\cdot (EF+DF+DE)$. Also we have$2S_{ABC}=r\cdot (AB+BC+AC)$,which $r$ is the radiu of the incircle of triangle $ABC$. It's well known that $R\ge 2r$ so we get $AB+BC+AC\ge 2(DE+DF+EF)$.