Clearly, the desired inequality holds if either $a = b$ or $a = c$, in which case a term on the LHS is equal to that on the RHS. If, say, $a = c$, equality holds iff $a^2 + c^2 = 4\Delta = 2ac \sin B$, i.e., $\sin B = 1$, which means that $ABC$ is right-angled with the right angle at $B$. If $a = b$, equality holds iff $ABC$ is right-angled with the right angle at $C$.

Suppose $b = c$. Then the inequality holds iff $4(a^2 + b^2 - 4\Delta) \ge (2b^2 - 4\Delta)$, equivalently, $2a^2 + b^2 \ge 6\Delta$, or $4a^4 + 4a^2b^2 + b^4 \ge 36\Delta^2$. Heron's formula, $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$, can be rewritten as
$$16\Delta^2 = (a+b+c)(-a+b+c)(a-b+c)(a+b-c).$$
When $b = c$, the right hand side becomes $(4b^2 - a^2)a^2$, thus we want to show
$$4a^4 + 4a^2b^2 + b^4 \ge \frac{9}{4}(4a^2b^2 - a^4).$$
Simplifying this, the desired inequality holds iff
$$25a^4 + 4b^4 \ge 20a^2b^2, \text{ namely, } (5a^2 - 2b^2)^2 \ge 0,$$
which is true, with equality iff $5a^2 = 2b^2$.

For an alternative way to prove $2a^2 + b^2 \ge 6\Delta$, observe that $a = 2b \sin(A/2)$ in an isosceles triangle with $b = c$. Using $2\Delta = bc \sin A = b^2 \sin A$, the desired inequality becomes
$$8b^2 \sin^2 \frac{A}{2} + b^2 \ge 3b^2 \sin A.$$
Upon cancelling $b^2$ and using $2\sin^2(A/2) = 1 - \cos A$, this simplifies to
$$4(1 - \cos A) + 1 \ge 3\sin A \quad \text{or} \quad 5 \ge 3\sin A + 4\cos A.$$
The latter is true because the tangent to the unit circle at $(x, y) = (4/5, 3/5)$ has equation $5x + 3y = 4$ and all points $(\cos A, \sin A)$ on the unit circle are below this tangent.

Equality occurs only for the point of tangency, which is $(\cos A, \sin A) = (4/5, 3/5)$. As $a = 2b \sin(A/2)$ and $2 \sin^2(A/2) = 1 - \cos A$, we see now that equality occurs iff $a^2 = 4b^2 \sin^2(A/2) = 2b^2(1 - \cos A) = 2b^2/5$.