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Question Number 109461 by 150505R last updated on 23/Aug/20

Answered by 1549442205PVT last updated on 24/Aug/20

$$\mathrm{Applying}\mathrm{the}\mathrm{inequality}\mathrm{AM}-\mathrm{GM}\mathrm{we}$$$$\mathrm{have}:{\left(2\mathrm{a}\right)}^2={(\mathrm{a}.\tan \alpha +\sqrt{{\mathrm{a}}^2-1}.\tan \beta +\sqrt{{\mathrm{a}}^2+1}.\tan \gamma )}^2$$$$⩽[({\mathrm{a}}^2+{\left(\sqrt{{\mathrm{a}}^2-1}\right)}^2+{\left(\sqrt{{\mathrm{a}}^2+1}\right)}^2]({\tan }^2\alpha +{\tan }^2\beta +{\tan }^2\gamma )$$$$=\left({3\mathrm{a}}^2\right)({\tan }^2\alpha +{\tan }^2\beta +{\tan }^2\gamma )$$$$\Rightarrow {\tan }^2\alpha +{\tan }^2\beta +{\tan }^2\gamma ⩾\frac{{4\mathrm{a}}^2}{{3\mathrm{a}}^2}=\frac{4}{3}$$$$\mathrm{The}\mathrm{equlity}\mathrm{ocurrs}\mathrm{if}\mathrm{and}\mathrm{only}\mathrm{if}$$$$\{\begin{array}{l}\frac{\mathrm{a}}{\tan \alpha }=\frac{\sqrt{{\mathrm{a}}^2-1}}{\tan \beta }=\frac{\sqrt{{\mathrm{a}}^2+1}}{\tan \gamma }\left(1\right)\\ {\tan }^2\alpha +{\tan }^2\beta +{\tan }^2\gamma =\frac{4}{3}\left(2\right)\end{array}$$$$\Rightarrow \frac{{\mathrm{a}}^2}{{\tan }^2\alpha }=\frac{{\mathrm{a}}^2-1}{{\tan }^2\beta }=\frac{{\mathrm{a}}^2+1}{{\tan }^2\gamma }=\frac{{3\mathrm{a}}^2}{4/3}$$$$\iff \{\begin{array}{l}{\tan }^2\alpha =\frac{4}{9}\\ {\tan }^2\beta =\frac{4({\mathrm{a}}^2-1)}{{9\mathrm{a}}^2}\\ {\tan }^2\gamma =\frac{4({\mathrm{a}}^2+1)}{{9\mathrm{a}}^2}\end{array}$$$$\iff \{\begin{array}{l}\tan \alpha =\pm \frac{2}{3}\\ \tan \beta =\pm \frac{2}{3\mathrm{a}}\sqrt{{\mathrm{a}}^2-1}\\ \tan \gamma =\pm \frac{2}{3}\sqrt{{\mathrm{a}}^2+1}\end{array}$$$$\mathrm{Thus},\mathrm{P}=3({\tan }^2\alpha +{\tan }^2\beta +{\tan }^2\gamma )$$$$\mathrm{has}\mathrm{the}\mathrm{least}\mathrm{value}\mathrm{equal}\mathrm{to}4\mathrm{when}$$$$(\mathit{\alpha },\mathit{\beta },\mathit{\gamma })\in \{{\mathbf{tan}}^{-1}(\pm \frac{2}{3})+\mathbf{k}\mathit{\pi },{\mathbf{tan}}^{-1}(\pm \frac{2}{3\mathbf{a}}\sqrt{{\mathbf{a}}^2-1})+\mathbf{k}\mathit{\pi },{\mathbf{tan}}^{-1}(\pm \frac{2}{3\mathbf{a}}\sqrt{{\mathbf{a}}^2+1})+\mathbf{k}\mathit{\pi }\}$$

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