inequality in geometry 
Given triangle $ABC$ has $BC=a$, $CA=b$, $AB=c$ and  $M$ is a point of the triangle plane. Prove that:
  \begin{align*}
  \cos \dfrac{A}{2} \cdot MA+\cos \dfrac{B}{2} \cdot MB+\cos \dfrac{C}{2} \cdot MC \geq \dfrac{a+b+c}{2}.
 \end{align*}

My attempts: 
\begin{eqnarray*}
   a \cdot \overrightarrow{IA}+b \cdot \overrightarrow{IB}+c \cdot \overrightarrow{IC}=\overrightarrow{0} \Rightarrow \dfrac{\cos \dfrac{A}{2}}{IA}\overrightarrow{IA}+\dfrac{\cos \dfrac{B}{2}}{IB}\overrightarrow{IB}+\dfrac{\cos \dfrac{C}{2}}{IC}\overrightarrow{IC}=\overrightarrow{0}??????
  \end{eqnarray*}
with $I$ is center of inscribed circle of triangle $ABC$.
\begin{eqnarray*}
   \cos \dfrac{A}{2} \cdot MA=\dfrac{\cos \dfrac{A}{2}}{IA} \cdot MA \cdot IA\geq \dfrac{\cos \dfrac{A}{2}}{IA} \cdot \overrightarrow{MA} \cdot \overrightarrow{IA}.
  \end{eqnarray*}
The same as,
\begin{eqnarray*}
   \cos \dfrac{B}{2} \cdot MB &\geq& \dfrac{\cos \dfrac{B}{2}}{IB} \cdot \overrightarrow{MB} \cdot \overrightarrow{IB}\\
 \cos \dfrac{C}{2} \cdot MC &\geq& \dfrac{\cos \dfrac{C}{2}}{IC} \cdot \overrightarrow{MC} \cdot \overrightarrow{IC}.
  \end{eqnarray*}
I am stuck here.

Here is my proof, based on the main result that if $DEF$ is the pedal triangle of $M$ then we have: $$\cos\left(\frac{A}{2}\right)\cdot MA\ge \frac{AE+AF}{2}\qquad (1)$$
In short, we prove the above inequality by using the formula for $\sin\alpha -\sin\beta$ to show that $$\cos\left(\frac{A}{2}\right) \cdot MA =  \dfrac{MF-ME}{2\sin\left(\dfrac{\widehat{MAF}-\widehat{MAE}}{2}\right)}\quad (2)$$
Then we use the following lemma: If $ABCD$ is a cyclic quadrilateral and $\widehat{B}=\widehat{D}=90^{\circ}$ then $$\dfrac{DA-DC}{BA+BC}\ge \sin\left(\dfrac{\widehat{ABD}-\widehat{CBD}}{2}\right) \qquad (3)$$
$(1)$ follows from $(2)$ and $(3)$, which gives us the original inequality
