<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] In the standard notation we need to prove that:
$$\sum_{cyc}\frac{a}{\dfrac{2S}{a}}\geq2\sum_{cyc}\sqrt{\frac{1-\frac{b^2+c^2-a^2}{2bc}}{1+\frac{b^2+c^2-a^2}{2bc}}}$$ or
$$\sum_{cyc}a^2\geq4S\sum_{cyc}\sqrt{\frac{(a+b-c)(a+c-b)}{(b+c-a)(a+b+c)}}$$ or
$$\sum_{cyc}a^2\geq\sum_{cyc}(a^2-(b-c)^2),$$ which is obvious.
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