Ireland

Denote by $D$ the mid-point of $BC$. We offer two ways of doing part (a).

First way: Continue the line segment $AD$ through $D$ to the point $A^{\prime }$ chosen so that $|A^{\prime }D|=|DA|=m_a$. Consider the triangles $A^{\prime }DC$ and $ADB$. Note that $\angle A^{\prime }DC=\angle ADB$, $|BD|=|DC|$ and $|A^{\prime }D|=|AD|$, by construction.



Hence, these triangles are congruent and so, in particular, $|A^{\prime }C|=|AB|$. Now consider the triangle $A^{\prime }CA$, and apply the triangle inequality to infer that $$2m_a=|A^{\prime }D|+|DA|=|A^{\prime }A|<|CA|+|A^{\prime }C|=b+c.$$

Second way: Two applications of the Cosine Rule tell us that $$\begin{gathered} a{m}_a\cos \angle BDA=m_a^2+{\left(\frac{a}{2}\right)}^2-c^2\text{and} \\ -a{m}_a\cos \angle BDA=a{m}_a\cos \angle CDA=m_a^2+{\left(\frac{a}{2}\right)}^2-b^2. \end{gathered}$$ Thus, eliminating $\cos \angle BDA$, we see that $$4m_a^2=2(b^2+c^2)-a^2.$$ Hence $2m_a<b+c$ iff $$2(b^2+c^2)-a^2<b^2+2bc+c^2⟺b^2+c^2-a^2<2bc⟺\cos (\angle BAC)<1,$$ which is true. In like manner, $2m_b<c+a$, $2m_c<a+b$, and so $$2m_a+2m_b+2m_c<(a+b)+(b+c)+(c+a)=2(a+b+c),$$ i.e. $m_a+m_b+m_c<a+b+c$. This completes the proof of part (a).

b. The numbers $2$, $4$, $5$ are the side lengths of a triangle $ABC$ with $a=4$, $b=2$, $c=5$ for which $$m_a^2=\frac{2(b^2+c^2)-a^2}{4}=\frac{21}{2}>10=bc.$$ Hence (i).

(ii) occurs in any triangle in which $b=c$, because, in such a case, $$bc=b^2=m_a^2+{\left(\frac{a}{2}\right)}^2>m_a^2.$$