Problem 1434

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.
$$-am_a \cos \theta = m_a^2 + \frac{a^2}{4} - b^2.$$
Hence $4m_a^2 = 2(b^2 + c^2) - a^2$ and so
$$\begin{aligned} 4m_a^2 &= b^2 + c^2 + 2bc \cos A \\ &= b^2 + c^2 - 2bc \cos(B + C) \\ &= (b \sin B + c \sin C)^2 + (b \cos B - c \cos C)^2. \end{aligned}$$
It follows that
$$2m_a \ge b \sin B + c \sin C,$$
with equality iff $b \cos B = c \cos C$, equivalently, iff
$$b^2(c^2 + a^2 - b^2) = c^2(a^2 + b^2 - c^2) \iff (b^2 - c^2)(a^2 - b^2 - c^2) = 0.$$
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
$$2m_b \ge c \sin C + a \sin A, \quad \text{and} \quad 2m_c \ge a \sin A + b \sin B,$$
whence adding these inequalities we deduce the stated result. Moreover, the inequality is strict unless $a = b = c$.