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