Question

Show ${\tan }^2\left(A/2\right)+{\tan }^2\left(B/2\right)+{\tan }^2\left(C/2\right)\ge 1$ when $A+B+C=\pi$ Hence the minimum value of $\sum {\tan }^2\left(A/2\right)$ is $1$.

Answer 1

We know that when $A+B+C=\pi$, then
$\tan \left(A/2\right)\tan \left(B/2\right)+\tan \left(B/2\right)\tan \left(C/2\right)+\tan \left(C/2\right)\tan \left(A/2\right)=1$ .$\left(1\right)$
or $yz+zx+xy=1$, where $x=\tan \left(A/2\right)$ etc.
Now we know that $(x-y{)}^2+(y-z{)}^2+(z-x{)}^2\ge 0$
or $2\sum x^2-2\sum xy\ge 0$
or $\sum x^2\ge \sum xy$ or $\sum x^2\ge 1$
$\because \sum xy=1$
or ${\tan }^2\left(A/2\right)+{\tan }^2\left(B/2\right)+{\tan }^2\left(C/2\right)\ge 1$
Note: For another method see the chapter on inequalities where we have proved that $x^2+y^2+z^2\ge xy+yz+zx=1\therefore x^2+y^2+z^2\ge 1$.