Question. In an acute-angle triangle $ABC$, what is the minimum value of $5\tan A+2\tan B+ \tan C$?
I have a form :
Given $5\tan A+2\tan B+\tan C$ in a $\triangle ABC$ and $A<90,B<90,C<90$. The function $f(A,B)=5\tan A+2\tan B-\tan(A+B)$ has an extremum if $$ \begin{align} \frac{\partial f(A,B)}{\partial A} &= 0 \\ \frac{\partial f(A,B)}{\partial B} &= 0. \end{align} $$ This gives us $$ \begin{align} \frac{5}{\cos^{2}A} &=\frac{1}{\cos^{2}(A+B)} \\ \frac{2}{\cos^{2}B} &=\frac{1}{\cos^{2}(A+B)}, \end{align} $$ or $$ \begin{align} 2\cos^{2}A &= 5\cos^{2}B \\ 2\cos^{2}(A+B) &=\cos^{2}B. \end{align} $$ Since $A<90,B<90,C<90$, we write $$ \sqrt{2}\cos A=\sqrt{5}\cos B \quad (1) $$ and $$ -\sqrt 2 \cos C=\sqrt 2 \cos(A+B)=-\cos B \quad (2). $$ From (2): $\sqrt{2}\cos A\cos B-\sqrt{2}\sin A\sin B=\cos B$. Using (1): $$ \sqrt{5}\cos^{2}B-\sqrt{2}\sin A\sin B=-\cos B, $$ $$ \sqrt{5}\cos^{2}B+\cos B=\sqrt{2}\sin A\sin B. $$ $$ (\sqrt{5}\cos^{2}B+\cos B)^{2}=2\sin^{2}A\sin^{2}B. $$ $$ 5\cos^{4}B+2\sqrt{5}\cos^{3}B+\cos^{2}B=(2-5\cos^{2}B)(1-\cos^{2}B). $$ $$ \sqrt{5}\cos^{3}B+4\cos^{2}B-1=0. $$ $$ \begin{align} \cos B &= \frac{1}{\sqrt{5}} \\ \cos A &= \frac{1}{\sqrt{2}}. \end{align} $$
Conclusion: $\tan A = 1$ and $\tan B = 2$ and $\tan C = -\tan(A+B)=3$.
There is another way to solve without calculation?
