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.

Use the sine rule and the area formuli of the triangle $\frac12 h_a a = \frac12 bc\sin\alpha$, $\frac12 h_b b = \frac12 ca\sin\beta$, $\frac12 h_c c = \frac12 ab\sin\gamma$ to express
$$\frac{a}{h_a} + \frac{b}{h_b} + \frac{c}{h_c} = \frac{\sin^2\alpha +\sin^2\beta+\sin^2\gamma}{\sin\alpha \sin\beta\sin\gamma}\tag 1$$
Then, evaluate
$$\sin^2\alpha- 2\sin\alpha \sin\beta\sin\gamma\cdot\tan\frac{\alpha}{2}$$
$$=4\sin^2\frac{\alpha}2\cos^2\frac{\alpha}2-4\sin^2\frac{\alpha}2\sin\beta\sin\gamma
=2\sin^2\frac{\alpha}2\left(2\cos^2\frac{\alpha}2-2\sin\beta\sin\gamma\right)$$
$$=2\sin^2\frac{\alpha}2\left(1+\cos\alpha-\cos(\beta-\gamma)+\cos(\beta+\gamma)\right)
= 4\sin^2\frac{\alpha}2\sin^2\frac{\beta-\gamma}2\tag 2\ge 0$$
and, similarly,
$$\sin^2\beta-2\sin\alpha\sin\beta\sin\gamma\cdot\tan\frac{\beta}{2} =4\sin^2\frac{\beta}2\sin^2\frac{\alpha-\gamma}2\ge 0 \tag 3$$
$$\sin^2\gamma-2\sin\alpha \sin\beta\sin\gamma\cdot\tan\frac{\gamma}{2}=4\sin^2\frac{\gamma}2\sin^2\frac{\alpha-\beta}2 \ge 0\tag 4$$
As a result, $(2)+(3)+(4)$ leads to
$$\sin^2\alpha +\sin^2\beta+\sin^2\gamma\ge 
2\sin\alpha \sin\beta\sin\gamma\left(\tan\frac{\alpha}{2}+\tan\frac{\beta}{2}+\tan\frac{\gamma}{2}\right)$$
Substitute the inequality into $(1)$ to obtain,
$$\frac{a}{h_a} + \frac{b}{h_b} + \frac{c}{h_c} \ge 2\left(\tan\frac{\alpha}{2}+\tan\frac{\beta}{2}+\tan\frac{\gamma}{2}\right)$$
