Problem 1640

1. Let $D$ be the point on $BC$ such that $AD$ is the angle bisector of $\angle BAC$. Also, let $E$ and $F$ be points on $AB$ and $AC$ respectively such that $DE \perp AB$ and $DF \perp AC$. Clearly, $AEDF$ lie on the circle with diameter $AD$. So, $EF = AD \sin A$.

2. By the Law of Sines in triangle $ABD$, we have:
$$AD \sin \left(\frac{A}{2}\right) = BD \sin B$$
Therefore,
$$AD \sin A = 2 BD \sin B \cos \left(\frac{A}{2}\right) = \frac{2ac}{b + c} \sin B \cos \left(\frac{A}{2}\right)$$
Hence,
$$l_a = \frac{2ac}{b + c} \sin B \cos \left(\frac{A}{2}\right)$$

3. Multiplying cyclically, we get:
$$l_a l_b l_c = \frac{8a^2 b^2 c^2}{(a + b)(b + c)(c + a)} \sin A \sin B \sin C \cos \left(\frac{A}{2}\right) \cos \left(\frac{B}{2}\right) \cos \left(\frac{C}{2}\right)$$

4. Therefore,
$$\frac{l_a l_b l_c}{l^3} = \frac{8a^2 b^2 c^2}{(a + b)(b + c)(c + a)(a + b + c)^3} \sin A \sin B \sin C \cos \left(\frac{A}{2}\right) \cos \left(\frac{B}{2}\right) \cos \left(\frac{C}{2}\right)$$

5. By the AM-GM inequality, we have:
$$\frac{8a^2 b^2 c^2}{(a + b)(b + c)(c + a)(a + b + c)^3} \leq \frac{1}{27}$$

6. Consider the function $f(x) = \sin x \cos \left(\frac{x}{2}\right)$. It is easy to see that $f''(x) < 0$ for $0 < x < \frac{\pi}{2}$. Thus, by Jensen's inequality for concave functions:
$$f(A) f(B) f(C) \leq \left(\frac{f(A) + f(B) + f(C)}{3}\right)^3 \leq \left(f\left(\frac{A + B + C}{3}\right)\right)^3 = \left(f\left(\frac{\pi}{3}\right)\right)^3 = \left(\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2}\right)^3 = \left(\frac{3}{4}\right)^3 = \frac{27}{64}$$

7. Hence,
$$\frac{8a^2 b^2 c^2}{(a + b)(b + c)(c + a)(a + b + c)^3} \sin A \sin B \sin C \cos \left(\frac{A}{2}\right) \cos \left(\frac{B}{2}\right) \cos \left(\frac{C}{2}\right) \leq \frac{1}{64}$$

$\blacksquare$