Question

If $A,B,C$ are angles of a triangle, then the minimum value of ${\tan }^2\frac{A}{2}+{\tan }^2\frac{B}{2}+{\tan }^2\frac{C}{2}$ is

Answer 1

Since, $A+B+C=\pi ,$ therefore,

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

$\Rightarrow xy+yz+zx=1$ ...(1)

where $x=\tan \frac{A}{2},y=\tan \frac{B}{2},z=\tan \frac{C}{2}$

We know that ${(x-y)}^2+{(y-z)}^2+{(z-x)}^2\ge 0$

$\Rightarrow 2\sum x^2>2\sum xy\Rightarrow \sum x^2>\sum xy$

$\Rightarrow \sum x^2\ge 1$ $[\because \sum xy=1$ (from(1))$]$

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

Thus, the minimum value of

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