Trigonometric Triangle Equality

Question

$A, B, C$ are the angles of a triangle then $tan^2(A/2)+tan^2(B/2)+tan^2(C/2)$ is always greater than what integral value.

Answer 1

Given

$$ \tan^2(A/2) + \tan^2(B/2) + \tan^2(C/2) \ge K $$

Question is: find $K$.

Write

$$ Q(x,y,z) = \tan^2(x) + \tan^2(y) + \tan^2(z), $$

such that

$$ x+y+z = P, $$

where $P$ is a constant. Then

$$ dQ = 2 \frac{\tan(x)}{\cos^2(x)} d x + 2 \frac{\tan(y)}{\cos^2(y)} dy + 2 \frac{\tan(z)}{\cos^2(z)} dz $$

But as $x+y+z=P$, we also have

$$ dx+dy+dz = 0 $$

so we obtain

$$ dQ = 2 \left( \frac{\tan(x)}{\cos^2(x)} - \frac{\tan(z)}{\cos^2(z)} \right) d x + 2 \left( \frac{\tan(y)}{\cos^2(y)} - \frac{\tan(z)}{\cos^2(z)} \right) dy $$

We find an extreme value for

$$ \frac{\tan(x)}{\cos^2(x)} - \frac{\tan(z)}{\cos^2(z)} = 0 $$

and

$$ \frac{\tan(y)}{\cos^2(y)} - \frac{\tan(z)}{\cos^2(z)} $$

So

$$ x = y = z = P/3 $$

Given that

$$ A + B + C = 180^o, $$

we have

$$ x+y+z = 90^o $$

So minimum for

$$ x=y=z=30^o $$

So we obtain

$$ K = 3 \tan^2(30^o) = 1, $$

whence

$$ \tan^2(A/2) + \tan^2(B/2) + \tan^2(C/2) \ge 1 $$

Answer 2

Since $A,B,C$ are angles of a triangle, we have $0<\dfrac{A}{2},\dfrac{B}{2},\dfrac{C}{2}<\dfrac{\pi}{2}$. For this range of values, $\tan^2 x$ is a convex function. Hence, from Jensen's inequlaity,

$$\tan^2\left(\frac{1}{3}\frac{A}{2}+\frac{1}{3}\frac{B}{2}+\frac{1}{3}\frac{C}{2}\right) \leq \frac{1}{3}\tan^2\frac{A}{2}+\frac{1}{3}\tan^2\frac{B}{2}+\frac{1}{3}\tan^2\frac{C}{2}$$ $$\tan^2\left(\frac{A+B+C}{6}\right) \leq \frac{1}{3}\left(\tan^2\frac{A}{2}+\tan^2\frac{B}{2}+\tan^2\frac{C}{2}\right)$$ Since $A+B+C=\pi$, hence

$$\tan^2\frac{A}{2}+\tan^2\frac{B}{2}+\tan^2\frac{C}{2} \geq 1$$

$\blacksquare$

Answer 3

Here $A,B,C$ are the angles of a $\triangle.$ So Here $$\displaystyle 0<\frac{A}{2},\frac{B}{2},\frac{C}{2}<\frac{\pi}{2}$$.

Now Using $\bf{A.M\geq G.M}\;,$ We Get

So we get $$\displaystyle \frac{\tan^2 \left(\frac{A}{2}\right)+\tan^2 \left(\frac{B}{2}\right)}{2}\geq \sqrt{\tan^2 \left(\frac{A}{2}\right)\cdot \tan^2 \left(\frac{B}{2}\right)} = \tan\left(\frac{A}{2}\right)\cdot \tan \left(\frac{B}{2}\right)\color{red}\checkmark$$

Similarly $$\displaystyle \frac{\tan^2 \left(\frac{B}{2}\right)+\tan^2 \left(\frac{C}{2}\right)}{2}\geq \sqrt{\tan^2 \left(\frac{B}{2}\right)\cdot \tan^2 \left(\frac{C}{2}\right)} = \tan \left(\frac{B}{2}\right)\cdot \tan \left(\frac{C}{2}\right)\color{red}\checkmark$$

Similarly $$\displaystyle \frac{\tan^2 \left(\frac{C}{2}\right)+\tan^2 \left(\frac{A}{2}\right)}{2}\geq \sqrt{\tan^2 \left(\frac{C}{2}\right)\cdot \tan^2 \left(\frac{A}{2}\right)} = \tan \left(\frac{C}{2}\right)\cdot \tan \left(\frac{A}{2}\right)\color{red}\checkmark$$

Now Add all Three, We Get

$$\displaystyle \tan^2\left(\frac{A}{2}\right)+\tan^2\left(\frac{B}{2}\right)+\tan^2\left(\frac{C}{2}\right)\geq \tan\left(\frac{A}{2}\right)\cdot \tan\left(\frac{B}{2}\right)+\tan\left(\frac{B}{2}\right)\cdot \tan\left(\frac{C}{2}\right)+\tan\left(\frac{C}{2}\right)\cdot \tan\left(\frac{A}{2}\right)\color{blue}\checkmark\color{blue}\checkmark$$

Now Here $$\displaystyle \frac{A}{2}+\frac{B}{2}=\frac{C}{2}\Rightarrow \tan\left(\frac{A}{2}+\frac{B}{2}\right)=\tan \frac{C}{2}$$

So We Get $$\displaystyle \tan\left(\frac{A}{2}\right)\cdot \tan\left(\frac{B}{2}\right)+\tan\left(\frac{B}{2}\right)\cdot \tan\left(\frac{C}{2}\right)+\tan\left(\frac{C}{2}\right)\cdot \tan\left(\frac{A}{2}\right)=1$$

Put into $\color{blue}\checkmark\color{blue}\checkmark\;,$ We Get

$$\displaystyle \tan^2\left(\frac{A}{2}\right)+\tan^2\left(\frac{B}{2}\right)+\tan^2\left(\frac{C}{2}\right)\geq 1$$

Answer 4

Minimum value is achieved when symmetric in $ A,B,C,a,b,c $ equilateral triangle case when they are equal, i.e., $ A = B = C = \pi/3$, which gives to the expression a minimum value

$$ 1/3 + 1/3 + 1/3 = 1.$$