\[ \sum \tan\frac{A}{2}\tan\frac{B}{2} = 1 \]
Denotes \((x, y, z) = (A/2, B/2, C/2)\), we can convert the expressoin about the angles inside triangle into the algebraic problem with limit \(\displaystyle \sum xy = 1\)
Example. Prove in \(\triangle ABC\) \[ \displaystyle \sum \sin\frac{A}{2}\le \frac{3}{2} \] Proof. \[ \begin{align} \sin\frac{A}{2} &= \frac{\sqrt{\sin^2\dfrac{A}{2}}}{\sin^2\dfrac{A}{2} + \cos^2\dfrac{A}{2}} \\ &= \frac{\sqrt{\tan^2\dfrac{A}{2}}}{\tan^2\dfrac{A}{2} + 1} \\ &= \sqrt{\frac{x^2}{x^2+xy+yz+zx}} \\ &= \sqrt{\frac{x}{x+y}}\sqrt{\frac{x}{x+z}} \\ \end{align} \] Hence \[ \displaystyle \sum \sin\frac{A}{2} = \sqrt{\frac{x}{x+y}}\sqrt{\frac{x}{x+z}} \le \frac{1}{2}\sum \left( \frac{x}{x+y} + \frac{x}{x+z}\right) = \frac{3}{2} \]
Example. Prove in \(\triangle ABC\) \[ \sum \tan\frac{A}{2} \prod\tan\frac{A}{2} \le \frac{1}{3} \] Proof. \[ \begin{align} \sum \tan\frac{A}{2} \prod\tan\frac{A}{2} &= \sum(xy+yz) \\ &\le \frac{1}{3}\sum(x^2y^2+2xy\cdot yz) \\ &= \frac{1}{3} \left(\sum xy \right)^2 = \frac{1}{3} \end{align} \]