For △ABC, prove $\frac a{h_a} + \frac b{h_b} + \frac c{h_c} \ge 2 (\tan\frac{\alpha}2+ \tan\frac{\beta}2 + \tan\frac{\gamma}2)$

Question

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.

Answer 1

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,

$$\frac{a}{h_a} + \frac{b}{h_b} + \frac{c}{h_c} \ge 2\left(\tan\frac{\alpha}{2}+\tan\frac{\beta}{2}+\tan\frac{\gamma}{2}\right)$$

Answer 2

Using the double-angle and half-angle trigonometric identities, we have that $$2 \cdot \sum_{cyc}\tan\frac{\beta}{2} = 2 \cdot \sum_{cyc}\frac{\sin\beta}{\cos^2\dfrac{\beta}{2}} = \frac{1}{R} \cdot \sum_{cyc}\frac{b}{\cos\beta + 1} = \frac{2abc}{R} \cdot \sum_{cyc}\frac{1}{(c + a)^2 - b^2}$$

$$ = \frac{8 \cdot [ABC]}{a + b + c} \cdot \sum_{cyc}\frac{1}{c + a - b} = \frac{1}{2 \cdot [ABC]}\cdot \sum_{cyc}(a + b - c)(b + c - a)$$

$$ = \frac{1}{2\cdot [ABC]} \cdot \sum_{cyc}[b^2 - (c - a)^2] \le \frac{a^2 + b^2 + c^2}{2 \cdot [ABC]} = \sum_{cyc}\frac{b}{h_b}$$

Answer 3

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.