find maximum and minimum of : $\tan (a)\tan (b)\tan (c)$

Question

find maximum of :

$1)$ $\tan (a)\tan (b)\tan (c)$

where : $a+b+c=\frac{π}{3}$

$a,b,c\in [0,\frac{π}{3}[$

and minimum of :

$2)$ $\tan (a)\tan (b)\tan (c)$

if : $a+b+c=2\frac{π}{3}$ , $a,b,c\in [\frac{π}{3},\frac{π}{2}[$

my attempt for maximum :

drap $f(x)=\tan x$ then $f"(x)=2\sec^{3} x\sin x>0$ so $f$ is convex function by Jensen inequality we obtaine :

$\tan a\tan b\tan c≤(\frac{\tan a\tan b\tan c}{3})^{3}$

$≤\frac{f(\frac{a+b+c}{3})}{9}$

$≤\frac{\tan \frac{π}{9}}{9}$

is my work correct ?

and what about of minimum ?

If any one have another ideas let we see!

Answer 1

You can reduce your problem to a problem in two variables: $$f(a,b)=\tan(a)\tan(b)\tan\left(\frac{\pi}{3}-a-b\right)$$ and computing the partial derivaties. $$\frac{\partial f(a,b)}{\partial a}=\sec ^2(a) \tan (b) \cot \left(a+b+\frac{\pi }{6}\right)-\tan (a) \tan (b) \csc ^2\left(a+b+\frac{\pi }{6}\right)$$ $$\frac{\partial f(a,b)}{\partial b}=\tan (a) \sec ^2(b) \cot \left(a+b+\frac{\pi }{6}\right)-\tan (a) \tan (b) \csc ^2\left(a+b+\frac{\pi }{6}\right)$$

Answer 2

For second, you can use $f(x)=\ln(\tan(x))$ and $f''(x)=\dfrac{\tan^4(x)-1}{\tan^2(x)}\ge0$ for $x\in [\frac{π}{3},\frac{π}{2}]$ and $x=2\pi/3$

Answer 3

1. For $a=b=c=\frac{\pi}{9}$ we obtain a value $\tan^3\frac{\pi}{9}.$

We'll prove that it's a maximal value.

Indeed, we need to prove that $$\sum_{cyc}\ln\tan{a}\leq3\ln\tan\frac{\pi}{9},$$ which is true because $$3\ln\tan\frac{\pi}{9}-\sum_{cyc}\ln\tan{a}=\sum_{cyc}\left(\ln\tan\frac{\pi}{9}-\ln\tan{a}+\frac{2}{\sin\frac{2\pi}{9}}\left(a-\frac{\pi}{9}\right)\right)\geq0.$$ Let $f(a)=\ln\tan\frac{\pi}{9}-\ln\tan{a}+\frac{2}{\sin\frac{2\pi}{9}}\left(a-\frac{\pi}{9}\right).$

Thus, $$f'(a)=-\frac{1}{\sin{a}\cos{a}}+\frac{2}{\sin\frac{2\pi}{9}}=\frac{2(\sin2a-\sin40^{\circ})}{\sin2a\sin40^{\circ}},$$ which gives $a_{min}=20^{\circ}$ and $f(a)\geq f(20^{\circ})=0.$