Given $2\sin A + \sin B = 2\sin C$ in a triangle find minimum of $\frac{5}{\sin A} + \frac{9}{\sin C}$ Given $2\sin A + \sin B = 2\sin C $ where $A,B,C$ are angles in a triangle find minimum of $\frac{5}{\sin A} + \frac{9}{\sin C}$
so $\sin C - \sin A = \frac{ \sin B}{2}$ so $2\cos \frac{C+A}{2} \sin \frac{C-A}{2} = \frac{ \sin B}{2} = 2\sin \frac{B}{2} \sin \frac{C-A}{2} $  so $\sin \frac{C-A}{2} = \frac{1}{4}$
I feel like I am awfully close here, but how do I finish it?
 A: \begin{align*}
\sin B &= 2 (\sin C - \sin A) \\
&=4 \cos \left(\frac{A+C}{2}\right) \sin \left(\frac{C-A}{2}\right)
\end{align*}
Moreover
\begin{align*}
\sin B &= \sin(\pi-(A+C)) \\
&=2 \sin \left(\frac{A+C}{2}\right) \cos \left(\frac{A+C}{2}\right)
\end{align*}
Therefore
\begin{align*}
\sin \left(\frac{A+C}{2}\right) &= 2 \sin \left(\frac{C-A}{2}\right) \\
\sin \left(\frac{A}{2}\right)\cos \left(\frac{C}{2}\right)+\sin \left(\frac{C}{2}\right)\cos \left(\frac{A}{2}\right)&=2\sin \left(\frac{C}{2}\right)\cos \left(\frac{A}{2}\right)-2\sin \left(\frac{A}{2}\right)\cos \left(\frac{C}{2}\right) \\
3\tan \left(\frac{A}{2}\right)&=\tan \left(\frac{C}{2}\right)
\end{align*}
Let $c=\tan \left(\frac{C}{2}\right)$ and $a=\tan \left(\frac{A}{2}\right)$. We have both $a$ and $c$ in the interval $(0,\infty)$. Using $\sin(2x)=\frac{2\tan x}{1+\tan^2 x}$, we have
\begin{align*}
\frac{5}{\sin A} + \frac{9}{\sin C}&= 5\left(\frac{1+a^2}{2a}\right)+9\left(\frac{1+c^2}{2c}\right)\\
&= \frac{5}{2a}+\frac{5a}{2}+\frac{9}{2c}+\frac{9c}{2}\\
&=\frac{5}{2a}+\frac{5a}{2}+\frac{3}{2a}+\frac{27a}{2}\\
&=16a+\frac{4}{a}
\end{align*}
By the AM-GM inequality, $a=\frac{1}{2}$ minimizes this function in $(0,\infty)$ and the minimum value is $16$.
A: In terms of the sides and circumradius:
$$2\sin A + \sin B = 2\sin C\iff \dfrac{a}{R} + \dfrac{b}{2R} = \dfrac{c}{R}\iff 2c=b+2a.$$
Then, you can turn your expression into a function of two variable, say $a,b$, using the same Law of Sines and take derivatives, using the formula:
$$R = \dfrac{abc}{4S},$$
where $S$ is the area.
