Belarus — Selection and Training Session

One can prove the following
Lemma. If $\alpha$, $\beta$, $\gamma$ are angles of a triangle then
$$\cos \alpha + \cos \beta + \cos \gamma + \cos 2\alpha + \cos 2\beta + \cos 2\gamma \ge 0.$$
Now, if $a$, $b$, $c$ are the side lengths of the triangle, $\alpha$, $\beta$, $\gamma$ are corresponding angles and $S$ is the area of this triangle, then we have
$$S(H_1H_2H_3) = S \cdot (1 - \cos^2 \alpha - \cos^2 \beta - \cos^2 \gamma) = \frac{1}{2} S \cdot (-1 - \cos 2\alpha - \cos 2\beta - \cos 2\gamma);$$
and
$$S(L_1L_2L_3) = S \cdot \frac{2abc}{(a+b)(b+c)(c+a)} = S \cdot \frac{2 \sin \alpha \sin \beta \sin \gamma}{\prod_{cycl} (\sin \alpha + \sin \beta)} = S \cdot \frac{\sin \alpha \sin \beta \sin \gamma}{4 \prod_{cycl} (\cos \frac{\gamma}{2} \cos \frac{\alpha-\beta}{2})}.$$
Further,
$$\begin{aligned} \cos \alpha + \cos \beta + (\cos \gamma - 1) &= 2 \cos \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2} - 2 \sin^2 \frac{\gamma}{2} = \\ &= 2 \sin \frac{\gamma}{2} (\cos \frac{\alpha - \beta}{2} - \cos \frac{\alpha + \beta}{2}) = 4 \sin \frac{\alpha}{2} \sin \frac{\beta}{2} \sin \frac{\gamma}{2}. \end{aligned}$$
So
$$\begin{aligned} S(L_1L_2L_3) &= \frac{1}{2} S \cdot \frac{\cos \alpha + \cos \beta + (\cos \gamma - 1)}{\cos \frac{\alpha-\beta}{2} \cos \frac{\beta-\gamma}{2} \cos \frac{\gamma-\alpha}{2}} \ge \frac{1}{2} S \cdot (\cos \alpha + \cos \beta + (\cos \gamma - 1)) \ge \\ &\ge \frac{1}{2} S \cdot (-1 - \cos 2\alpha - \cos 2\beta - \cos 2\gamma) = S(H_1H_2H_3), \end{aligned}$$
as required.