SAMC

Denote by $a,b,c$ the length sides of triangle $ABC$. In triangle $ADM$ we have $$\begin{array}{c}D{M}^2=A{M}^2-A{D}^2=\frac{2\left(b^2+c^2\right)-a^2}{4}-\frac{4K^2}{a^2}\\ =\frac{2\left(b^2+c^2\right)-a^2}{4}-\frac{16K^2}{4a^2}\end{array}$$ $$\begin{array}{c}=\frac{2\left(b^2+c^2\right)-a^2}{4}-\frac{1}{4a^2}\left(2\sum a^2{b}^2-\sum a^4\right)\\ =\frac{1}{4a^2}\left(-2b^2{c}^2+b^4+c^4\right)={\left(\frac{b^2-c^2}{2a}\right)}^2\end{array}$$ Since $\hat{B}\ge 2\hat{C}>\hat{C}$, it follows $b>c$, hence $$\begin{array}{c}DM=\frac{b^2-c^2}{2a}\end{array}\text{\hspace{1em}(1)}$$ The inequality $DM\ge \frac{AB}{2}$ is equivalent to $\frac{b^2-c^2}{2a}\ge \frac{c}{2}$, that is $b^2\ge c^2+ac$. Using the cosine law, the last inequality becomes $a\ge 2c\cos B+c$, or $\sin A\ge 2\sin C\cos B+\sin C$. We can write $\sin (B+C)\ge 2\sin C\cos B+\sin C$, hence $\sin (B-C)\ge \sin C$, and we get $2\sin \frac{B-2C}{2}\cos \frac{B}{2}\ge 0$. This inequality is true because $\hat{B}\ge 2\hat{C}$, and we are done.