Let the area of the triangle be denoted by $A$. It is easy to see that $ah_a = bh_b = ch_c = 2A$. Hence the original proposition is equivalent to proving
$$a^4 + b^4 + c^4 \ge 16A^2. \quad (1)$$
Subtracting the right side of (1) from the left side, and using Heron's formula, we get
$$\begin{aligned} & a^4 + b^4 + c^4 - 16A^2 \\ = & a^4 + b^4 + c^4 - (a+b+c)(a+b-c)(a-b+c)(-a+b+c) \\ = & 2a^4 + 2b^4 + 2c^4 - 2a^2b^2 - 2a^2c^2 - 2b^2c^2 \\ = & (a^2 - b^2)^2 + (b^2 - c^2)^2 + (c^2 - a^2)^2 \ge 0. \end{aligned}$$
Thus (2) is proved, and the original proposition follows.