Ta có các góc là $A;\ B;\ C$ lập thành csc
$A+B+C=180^0$ mà $A+C =2B \Rightarrow 3B=180^0 \Rightarrow B=60^0 \Rightarrow A+C=120^0$
Vậy $\sin \dfrac{A}{2}+ \sin \dfrac{B}{2}+ \sin \dfrac{C}{2} =1+\dfrac{\sqrt 3}{3}$
$\Leftrightarrow \sin \dfrac{A}{2}+ \sin \dfrac{C}{2} =1+\dfrac{\sqrt 3}{3}-\sin 30^0 =\dfrac{1}{2}+\dfrac{\sqrt 3}{3}$
$\Leftrightarrow 2\sin \dfrac{A+C}{4}\cos \dfrac{A-C}{4}=\dfrac{1}{2}+\dfrac{\sqrt 3}{3}$
$\Leftrightarrow 2\sin 30^0 \cos \dfrac{A-C}{4}=\dfrac{1}{2}+\dfrac{\sqrt 3}{3}$
$\Leftrightarrow \cos \dfrac{A-C}{4}=\dfrac{1}{2}+\dfrac{\sqrt 3}{3}$ vô nghiệm