Irska

Let the median from $A$ meet the side $BC$ at $D$, so that $|BD|=|DC|=a/2$. We begin by deriving an expression for $m_a$. Let $\theta =\angle BDA$.



By the Cosine Rule, $2|BD||AD|\cos \theta =|AD{|}^2+|BD{|}^2-|AB{|}^2$, i.e. $$am{m}_a\cos \theta =m_a^2+\frac{a^2}{4}-c^2.$$ Similarly $2|CD||AD|\cos (\pi -\theta )=|AD{|}^2+|CD{|}^2-|AC{|}^2$, i.e. $$-a{m}_a\cos \theta =m_a^2+\frac{a^2}{4}-b^2.$$ Hence $4m_a^2=2(b^2+c^2)-a^2$ and so $$\begin{array}{rl}4m_a^2& =b^2+c^2+2bc\cos A\\ & =b^2+c^2-2bc\cos (B+C)\\ & =(b\sin B+c\sin C{)}^2+(b\cos B-c\cos C{)}^2.\end{array}$$ It follows that $$2m_a\ge b\sin B+c\sin C,$$ with equality iff $b\cos B=c\cos C$, equivalently, iff $$b^2(c^2+a^2-b^2)=c^2(a^2+b^2-c^2)⟺(b^2-c^2)(a^2-b^2-c^2)=0.$$ In other words, $$2m_a\ge b\sin B+c\sin C,$$ and there is equality iff either $b=c$ or $A$ is a right-angle. In the same way we see that $$2m_b\ge c\sin C+a\sin A,\text{and}2m_c\ge a\sin A+b\sin B,$$ whence adding these inequalities we deduce the stated result. Moreover, the inequality is strict unless $a=b=c$.