If $4\sin^2x + \csc^2x , a , \sin^2 y+ 4 \csc^2 y$ are in arithmetic progression, then find the minimum value of $6a$.
My Work:
Well just at a glance this problem seems to be related to AM-GM inequality. However whenever we apply am-gm we get the minimum value of $4\sin^2 x+\csc^2x$ and $\sin^2 y+4 \csc^2y$; but we do not get anything related to the arithmetic progression or anything related to $a$. So I need help as to any other way to attack this problem.
$a$ has to be the arithmetic mean of the terms before and after it so we can apply AM-GM:
\begin{align} 6a &= 3\left(4\sin^2 x + \frac1{\sin^2 x} + \sin^2 y + \frac4{\sin^2 y}\right) \\ &= 12\left(\sin^2 x+\frac1{\sin^2 y}\right) + 3\left(\sin^2 y+\frac1{\sin^2 x}\right)\\ &\ge 24\sqrt{\frac{\sin^2 x}{\sin^2 y}} + 6\sqrt{\frac{\sin^2 y}{\sin^2 x}}\\ &\ge 2\sqrt{24\sqrt{\frac{\sin^2 x}{\sin^2 y}} \cdot 6\sqrt{\frac{\sin^2 y}{\sin^2 x}}}\\ &= 2\sqrt{24 \cdot 6}\\ &= 24 \end{align} with equalities for $\sin^2 x = \sin^2 y = 1$ so $a = 4$ is indeed the minimum.
Alternatively, you can bound $$a =\frac{4\sin^2 x + \frac1{\sin^2 x} + \sin^2 y + \frac4{\sin^2 y}}2 \ge \sqrt{4\sin^2 x \cdot \frac1{\sin^2 x}} + \sqrt{\sin^2 y \cdot \frac4{\sin^2 y}} = 2+2 = 4$$ but it is not immediately visible how to attain the bound.