Solution:

Taking into consideration that the triangle $ABC$ is acute-angled, using the formulae
$$4 m_a^2 = 2 b^2 + 2 c^2 - a^2, \quad 4 m_b^2 = 2 a^2 + 2 c^2 - b^2, \quad 4 m_c^2 = 2 a^2 + 2 b^2 - c^2$$
and using the notations $x = b^2 + c^2 - a^2 > 0$, $y = a^2 + c^2 - b^2 > 0$, $z = a^2 + b^2 - c^2 > 0$, we shall prove the equivalent inequality
$$\frac{4x + y + z}{x} + \frac{4y + z + x}{y} + \frac{4z + x + y}{z} \geq 18$$
But
$$\frac{4x + y + z}{x} + \frac{4y + z + x}{y} + \frac{4z + x + y}{z} = 4\left(\frac{x}{x} + \frac{y}{y} + \frac{z}{z}\right) + \frac{y + z}{x} + \frac{z + x}{y} + \frac{x + y}{z} = 12 + \left(\frac{y}{x} + \frac{z}{x} + \frac{z}{y} + \frac{x}{y} + \frac{x}{z} + \frac{y}{z}\right)$$
By the AM-GM inequality,
$$\frac{y}{x} + \frac{z}{x} + \frac{z}{y} + \frac{x}{y} + \frac{x}{z} + \frac{y}{z} \geq 6$$
Therefore,
$$12 + 6 = 18$$
Equality holds if and only if $x = y = z$, that is, for the equilateral triangle.