$\tan{\frac{a}{2}}\cdot \tan{\frac{b}{2}}\cdot \tan{\frac{c}{2}}\leq \frac{1}{3\sqrt{3}}$, Where a,b,c are angles of triangle

Question

As in title $$\tan{\frac{a}{2}}\cdot \tan{\frac{b}{2}}\cdot \tan{\frac{c}{2}}\leq \frac{1}{3\sqrt{3}}$$whats more, is that this is acute triangle. I think it should be doable somehow with Jensen inequality, but thats the point where i am getting confused - how define necessary function, what to do further, any hints would be appreciated, thanks in advance

Answer 1

Method 1 - Jensen's inequality.

If an angle $\theta$ is acute, then $\frac{\theta}{2} \in (0,\frac{\pi}{4})$. Over the interval $(0,\frac{\pi}{4})$, we have

$$0 < \tan(x) < 1 \implies \frac{d^2}{dx^2} \log(\tan x) = \tan(x)^2 - \frac{1}{\tan(x)^2} < 0.$$

This means $\log(\tan x)$ is a strictly concave function there. Apply Jensen's inequality to $\frac{a}{2}, \frac{b}{2}, \frac{c}{2}$ and notice $a + b + c = \pi$, we get:

$$\begin{align}\log(\tan\frac{a}{2}) + \log(\tan\frac{b}{2}) + \log(\tan\frac{c}{2}) \le 3\log(\tan\frac{\pi}{6})\\ \implies \tan\frac{a}{2} \tan\frac{b}{2}\tan\frac{c}{2} \le (\tan\frac{\pi}{6})^3 = (\frac{1}{\sqrt{3}})^3 = \frac{1}{3\sqrt{3}} \end{align}$$

Method 2 - G.M. $\le$ A.M. and angle addition formula.

Recall the 3 angle addition formula for tangent:

$$\tan(\alpha + \beta + \gamma) = \frac{\tan\alpha + \tan\beta + \tan\gamma - \tan\alpha\tan\beta\tan\gamma} {1 - (\tan\alpha\tan\beta + \tan\beta\tan\gamma + \tan\gamma\tan\alpha)}$$

Notice $a + b + c = \pi \implies \tan(\frac{a}{2}+\frac{b}{2}+\frac{c}{2}) = \tan\frac{\pi}{2} = \infty$, we have:

$$\tan\frac{a}{2}\tan\frac{b}{2} + \tan\frac{b}{2}\tan\frac{c}{2} + \tan\frac{c}{2}\tan\frac{a}{2} = 1\tag{*}$$

Now $\tan\frac{a}{2}, \tan\frac{b}{2}, \tan\frac{c}{2} > 0$. Apply G.M. $\le$ A.M. to numbers on L.H.S of $(*)$, we get:

$$ \left(\tan\frac{a}{2}\tan\frac{b}{2}\tan\frac{c}{2}\right)^{\frac23} \le \frac13 \;\;\implies\;\;\tan\frac{a}{2}\tan\frac{b}{2}\tan\frac{c}{2} \le \frac{1}{3\sqrt{3}} $$