Let $a$, $b$ and $c$ be the lengths of the opposite sides of vertices $A$, $B$ and $C$, respectively. Also, let $D$ be the intersection of $BC$ and the angle bisector of $\angle A$. Put $p = BD$ and $q = CD$. Since $AD$ is the angle bisector of $\angle A$, we have $bp = cq$ and thus a little manipulation yields
$$p = \frac{ac}{b+c} \quad \text{and} \quad q = \frac{ab}{b+c}. \qquad (1)$$
Meanwhile, since $\cos(\angle ADB) + \cos(\angle ADC) = 0$, we have by the Cosine Law
$$\frac{x^2 + p^2 - c^2}{2px} + \frac{x^2 + q^2 - b^2}{2qx} = 0$$
where $x = AD$. This, together with (1), yields
$$x^2 = bc - pq = bc \left[ 1 - \left( \frac{a}{b+c} \right)^2 \right] = \frac{bc(b+c-a)}{(b+c)^2} \ell. \qquad (2)$$
Now, let $E$ and $F$ be the feet of perpendiculars drawn from $D$ to $AB$ and $AC$, respectively. Then, since $\square AEDF$ is inscribed in a circle, we have $\angle DEF = \angle DAF$. Accordingly, applying the Sine Law, we have
$$\frac{\ell_A}{\sin A} = \frac{\ell_A}{\sin(\angle EDF)} = \frac{DF}{\sin(\angle DEF)} = \frac{DF}{\sin(\angle DAF)} = x.$$
Thus, denoting the area of $\triangle ABC$ by $T$, we have by (2)
$$\ell_A = x \sin A = \frac{2xT}{bc} = \frac{2T}{(b+c)\sqrt{bc}} \sqrt{(b+c-a)\ell}.$$
Parallel arguments yield
$$\ell_B = \frac{2T}{(c+a)\sqrt{ca}} \sqrt{(c+a-b)\ell}$$

and
$$\ell_C = \frac{2T}{(a+b)\sqrt{ab}} \sqrt{(a+b-c)\ell}.$$
Hence, using Heron's formula, we obtain
$$\begin{aligned} \ell_A \ell_B \ell_C &= \frac{8T^3 \ell \sqrt{(a+b-c)(b+c-a)(c+a-b)\ell}}{abc(a+b)(b+c)(c+a)} \\ &= \frac{\ell^3 (a+b-c)^2 (b+c-a)^2 (c+a-b)^2}{8abc(a+b)(b+c)(c+a)}. \end{aligned}$$
Since $a + b \ge 2\sqrt{ab}$, $b + c \ge 2\sqrt{bc}$, $c + a \ge 2\sqrt{ca}$ by the AM-GM inequality, we deduce from the above that
$$\frac{\ell_A \ell_B \ell_C}{\ell^3} \le \frac{(a+b-c)^2 (b+c-a)^2 (c+a-b)^2}{64a^2b^2c^2}. \quad (3)$$
Moreover, since $a$, $b$ and $c$ are side lengths of a triangle, we have
$$\begin{aligned} 0 < (a+b-c)(c+a-b) &= a^2 - (b-c)^2 \le a^2, \\ 0 < (a+b-c)(b+c-a) &= b^2 - (a-c)^2 \le b^2, \\ 0 < (b+c-a)(c+a-b) &= c^2 - (a-b)^2 \le c^2, \end{aligned}$$
and thus
$$0 < (a-b+c)^2(b-a+c)^2(c-a+b)^2 \le a^2b^2c^2. \quad (4)$$
From inequalities (3) and (4) follows the required inequality. $\square$