2
$\begingroup$

What is the range of $\tan^2\frac A2+\tan^2\frac B2+\tan^2 \frac C2$ if $A,B,C$ are angles in a triangle?

For $\tan^2\frac A2+\tan^2\frac B2+\tan^2 \frac C2$, I know we have to apply the AM–GM inequality. So $$\begin{align}\frac12\left(\tan^2\frac A2+\tan^2\frac B2+\tan^2\frac C2\right)&\ge\sqrt{\tan^2\frac A2\tan^2\frac B2\tan^2\frac C2}\\ &\ge\tan\frac A2\tan\frac B2\tan\frac C2\\ \tan^2\frac A2+\tan^2\frac B2+\tan^2\frac C2&\ge2\tan\frac A2\tan\frac B2\tan\frac C2 \end{align}$$ After this, the LHS is of the required form, but the RHS is having half-angle terms. Then again, if $A+B+C=\pi$, $\tan\frac{A+B+C}2=\tan90^\circ=\infty$. Now I am stuck; how to solve further?

The given options are:

  1. $>1$
  2. $<1$
  3. $\ge1$
  4. $\le1$
$\endgroup$
2
$\begingroup$

Note $\tan^2(x)$ is convex, so by Jensen's inequality $$\sum_{cyc} \tan^2 \frac{A}2 \geqslant 3\tan^2 \frac{A+B+C}{2\cdot3} = 1$$

Further as $\tan(x)$ is unbounded, clearly we can have the sum unbounded by allowing one of the angles to approach $\pi$.

$\endgroup$
1
$\begingroup$

HINT:

As $\tan\left(\dfrac{A+B}2\right)=\tan\left(\dfrac{\pi- C}2\right)$

$$\implies\dfrac{\tan\dfrac A2+\tan\dfrac B2}{1-\tan\dfrac A2\tan\dfrac B2}=\dfrac1{\tan\dfrac C2}$$

$$\implies\tan\dfrac A2\tan\dfrac B2+\tan\dfrac B2\tan\dfrac C2+\tan\dfrac C2\tan\dfrac A2=1$$

Now $$\left(\tan\dfrac A2-\tan\dfrac B2\right)^2+\left(\tan\dfrac B2-\tan\dfrac C2\right)^2+\left(\tan\dfrac C2-\tan\dfrac A2\right)^2\ge0$$

$\endgroup$
0
$\begingroup$

Let $a+b-c=z$, $a+c-b=y$ and $b+c-a=x$.

Hence, by C-S $\sum\limits_{cyc}\tan^2\frac{\alpha}{2}=\sum\limits_{cyc}\frac{yz}{x(x+y+z)}=\sum\limits_{cyc}\frac{y^2z^2}{xyz(x+y+z)}\geq\frac{(xy+xz+yz)^2}{3xyz(x+y+z)}\geq1$.

The equality occurs for $x=y=z$, id est, for $a=b=c$.

In another hand, our sum $\rightarrow+\infty$ for $\alpha\rightarrow180^{\circ}$.

Id est, the answer is $[1,+\infty)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.