<s>[INST] Find the minimum value of 6a. 
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.
 [/INST] $a$ 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.
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