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 $$S(H_1{H}_2{H}_3)=S\cdot (1-{\cos }^2\alpha -{\cos }^2\beta -{\cos }^2\gamma )=\frac{1}{2}S\cdot (-1-\cos 2\alpha -\cos 2\beta -\cos 2\gamma );$$ and $$S(L_1{L}_2{L}_3)=S\cdot \frac{2abc}{(a+b)(b+c)(c+a)}=S\cdot \frac{2\sin \alpha \sin \beta \sin \gamma }{{\prod }_{cycl}(\sin \alpha +\sin \beta )}=S\cdot \frac{\sin \alpha \sin \beta \sin \gamma }{4{\prod }_{cycl}(\cos \frac{\gamma }{2}\cos \frac{\alpha -\beta }{2})}.$$ Further, $$\begin{array}{rl}\cos \alpha +\cos \beta +(\cos \gamma -1)& =2\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}-2{\sin }^2\frac{\gamma }{2}=\\ & =2\sin \frac{\gamma }{2}(\cos \frac{\alpha -\beta }{2}-\cos \frac{\alpha +\beta }{2})=4\sin \frac{\alpha }{2}\sin \frac{\beta }{2}\sin \frac{\gamma }{2}.\end{array}$$ So $$\begin{array}{rl}S(L_1{L}_2{L}_3)& =\frac{1}{2}S\cdot \frac{\cos \alpha +\cos \beta +(\cos \gamma -1)}{\cos \frac{\alpha -\beta }{2}\cos \frac{\beta -\gamma }{2}\cos \frac{\gamma -\alpha }{2}}\ge \frac{1}{2}S\cdot (\cos \alpha +\cos \beta +(\cos \gamma -1))\ge \\ & \ge \frac{1}{2}S\cdot (-1-\cos 2\alpha -\cos 2\beta -\cos 2\gamma )=S(H_1{H}_2{H}_3),\end{array}$$ as required.