Question

If $a>0$ and $A,B,C$ are variable angles of $△ABC,$ such that $a\tan A+2a\tan B+3a\tan C=7a,$ then the minimum integral value of ${\tan }^{2}A+{\tan }^{2}B+{\tan }^{2}C=$

Answer 1

Let two vectors, $\overrightarrow{P}=a\hat{i}+2a\hat{j}+3a\hat{k}$ and $\overrightarrow{Q}=\tan A\hat{i}+\tan B\hat{j}+\tan C\hat{k}$
If $\theta$ is the angle between $\overrightarrow{P}$ and $\overrightarrow{Q}$
then
$\cos \theta =\frac{\overrightarrow{P}\cdot \overrightarrow{Q}}{|\overrightarrow{P}||\overrightarrow{Q}|}=\frac{a\tan A+2a\tan B+3a\tan C}{\sqrt{14}a\sqrt{{\tan }^{2}A+{\tan }^{2}B+{\tan }^{2}C}}=\frac{7a}{\sqrt{14}a\sqrt{{\tan }^{2}A+{\tan }^{2}B+{\tan }^{2}C}}$
$\because {\cos }^2\theta \le 1\Rightarrow \frac{49}{14({\tan }^{2}A+{\tan }^{2}B+{\tan }^{2}C)}\le 1\Rightarrow {\tan }^{2}A+{\tan }^{2}B+{\tan }^{2}C\ge \frac{49}{14}$
$\therefore$ minimum integral value of ${\tan }^{2}A+{\tan }^{2}B+{\tan }^{2}C$ is $4.$