tangent inequality in triangle Let $a$, $b$ and $c$ be the measures of angles of a triangle (in radians).
It is asked to prove that 
$$\tan^2\left(\dfrac{\pi-a}{4}\right)+\tan^2\left(\dfrac{\pi-b}{4}\right)+\tan^2\left(\dfrac{\pi-c}{4}\right) \ge 1$$
When does equality occur ?
My try :
Letting $u:= \tan\left(\dfrac{\pi-a}{4}\right)$ and $v:= \tan\left(\dfrac{\pi-b}{4}\right)$ the inequality reduces to proving
$$u^2+v^2+\dfrac{(1-uv)^2}{(u+v)^2} \ge 1\quad\quad (*)$$ ($$u,v\in (0,1)$$)
( using $a+b+c=\pi$ and the formula for $\tan(x+y)$ and the fact that $\tan\left(\dfrac{\pi}{2}-x\right)=\dfrac{1}{\tan x}$ )
I'm having trouble in proving that last inequality.
Any suggestions are welcome.
Thanks.
Edit : is the following reasoning to prove the inequality (*) sound ?
(*) is obvious when $u^2 + v^2 \ge 1$ so we only have to deal with the case $u^2 + v^2 \le 1$ which we assume true in what follows.
(*) $\iff (u^2 + v^2)(u + v)^2 \ge (u+v)^2+(1-uv)^2$
Setting $x:= u^2 + v^2$ and $a:=uv$ we get
(*) $\iff x^2+(2a-1)x \ge a^2+4a-1$
Now some calculus $(x^2+(2a-1)x)' = 2x + 2a-1$
the function $\phi:x\mapsto x^2+(2a-1)x$ then has a minimum at $\dfrac 12 - a$ which is $\phi\left(\dfrac 12 - a\right) = -a^2+a-\dfrac 14$
It then suffices to have $-a^2+a-\dfrac 14 \ge a^2+4a-1$ 
This last ineq is equivalent to
$8a^2+12a-3 \le 0$ 
which, in turn, is equivalent to
$a \in \left[\dfrac{-6-\sqrt{60}}{8}, \dfrac{-6+\sqrt{60}}{8}\right]$
Recall that we're working under the assumption $u^2 +v^2 \le 1$, that yields in particular that $uv \le \dfrac 12$.
since $ \dfrac{-6-\sqrt{60}}{8} \le 0 \le a := uv \le \dfrac 12 \le \dfrac{-6+\sqrt{60}}{8}$, we're done.
Thanks for taking time to check the correctness of the above proof.
 A: Let $\dfrac{\pi-A}4=x$  etc.
$\implies4(x+y+z)=3\pi-\pi\iff x+y+z=\dfrac\pi2$
Now as $\tan x,\tan y,\tan z$ are real,
$$(\tan x-\tan y)^2+(\tan y-\tan z)^2+(\tan z-\tan x)^2\ge0$$
$$\implies\tan^2x+\tan^2y+\tan^2z\ge\tan x\tan y+\tan y\tan z+\tan z\tan x$$
Finally $$\tan(x+y)=\tan\left(\dfrac\pi2-z\right)$$
$$\iff\dfrac{\tan x+\tan y}{1-\tan x\tan y}=\dfrac1{\tan z}$$
Simplify
A: $$\left(\tan^2\frac{\pi-x}{4}\right)''=\frac{2-\sin\frac{x}{2}}{8\cos^4\frac{\pi-x}{4}}>0.$$
Thus, by Jensen
$$\sum_{cyc}\tan^2\frac{\pi-\alpha}{4}\geq3\tan^2\frac{\pi-\frac{\alpha+\beta+\gamma}{3}}{4}=1.$$
A: Using the inequality for concave upward graph of $y\,=({tan\,(x)})^2$.
That is $\frac{(\sum_{}^{}{tan}^2{\frac{180-A}{4})}}{3} \;$ $ \ge\,$ ${tan}^2(({\frac{(180-A)+(180-B)+(180-C)}{4}} )/3)$ =1/3
Hence $(\sum_{}^{}{tan}^2{\frac{180-A}{4}})$ $\ge\,1$
A: A different formulation:
Let
$\displaystyle \alpha = \frac{\pi - A}{4}$
$\displaystyle \beta = \frac{\pi - B}{4}$
$\displaystyle \gamma = \frac{\pi - C}{4}$
We need to find the minimum value of 
$\displaystyle \tan^2 \alpha + \tan^2 \beta + \tan^2 \gamma$ 
subject to 
$\displaystyle \alpha + \beta + \gamma = \frac{\pi}{2}$
Also $\displaystyle 0 \leq \alpha, \beta, \gamma \leq \frac{\pi}{4}$
Let 
$\displaystyle f(\alpha, \beta, \gamma) = \tan^2 \alpha + \tan^2 \beta + \tan^2 \gamma + \lambda \left(\alpha + \beta + \gamma - \frac{\pi}{2} \right)$
Now set 
$\displaystyle \frac{\partial f}{\partial \alpha} = \frac{\partial f}{\partial \beta} = \frac{\partial f}{\partial \gamma} = 0$
We immediately have 
$\displaystyle \alpha = \beta = \gamma = \frac{\pi}{6}$
and finally the minimum value of $f$ is 
$\displaystyle f \left( \frac{\pi}{6}, \frac{\pi}{6}, \frac{\pi}{6} \right) = 1$
