\because 0 <\theta < \pi

\therefore 0 <\dfrac{ \theta}{2} <\dfrac{ \pi}{2}

\sin \dfrac{ \theta}{2}>0, \cos\dfrac{ \theta}{2} >0

Using the inequality of arithmetic and geometric means

GM \leq AM

\sin \dfrac{\theta }{2} (1+\cos \theta )

=2\sin \dfrac{\theta }{2} \cos^2 \dfrac{\theta }{2}

=\sqrt{2\cdotp2\sin^2 \dfrac{\theta }{2} \cos^2 \dfrac{\theta }{2} \cos^2 \dfrac{\theta }{2} }

\leq \sqrt{2\cdotp \Bigg( \dfrac{ 2\sin^2 \dfrac{\theta }{2}+ \cos^2 \dfrac{\theta }{2} +\cos^2 \dfrac{\theta }{2} }{3} \Bigg) ^3 }

=\sqrt{\dfrac{16}{27} } =\dfrac{4\sqrt{3} }{9}