<s>[INST] For △ABC, prove $\frac a{h_a} + \frac b{h_b} + \frac c{h_c} \ge 2 (\tan\frac{\alpha}2+ \tan\frac{\beta}2 + \tan\frac{\gamma}2)$ 
Given $\triangle ABC$, (using the main parameters and notation), prove that $$ \frac{a}{h_a} + \frac{b}{h_b} + \frac{c}{h_c} \ge 2 \cdot \left(\tan\frac{\alpha}{2} + \tan\frac{\beta}{2} + \tan\frac{\gamma}{2}\right)$$

Great, another problem that already has a solution. We have many ways of expressing the area of a triangle, for example $$2 \cdot [ABC] = \frac{abc}{2R} = \sqrt{2(a + b + c) \cdot \sum_{cyc}\frac{c + a - b}{2}} = ah_a = bh_b = ch_c$$
The above equations are used in the solution I have provided below.
I would be greatly appreciated if you could come up with any other solutions.
 [/INST] Using the double-angle and half-angle trigonometric identities, we have that $$2 \cdot \sum_{cyc}\tan\frac{\beta}{2} = 2 \cdot \sum_{cyc}\frac{\sin\beta}{\cos^2\dfrac{\beta}{2}} = \frac{1}{R} \cdot \sum_{cyc}\frac{b}{\cos\beta + 1} = \frac{2abc}{R} \cdot \sum_{cyc}\frac{1}{(c + a)^2 - b^2}$$
$$ = \frac{8 \cdot [ABC]}{a + b + c} \cdot \sum_{cyc}\frac{1}{c + a - b} = \frac{1}{2 \cdot [ABC]}\cdot \sum_{cyc}(a + b - c)(b + c - a)$$
$$ = \frac{1}{2\cdot [ABC]} \cdot \sum_{cyc}[b^2 - (c - a)^2] \le \frac{a^2 + b^2 + c^2}{2 \cdot [ABC]} = \sum_{cyc}\frac{b}{h_b}$$
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