TITLE: AM-GM giving wrong result on applying to trigonometric functions
QUESTION [5 upvotes]: Question:

Find the range of $f(x)=\operatorname{cosec} ^2x+25\sec^2x$

My attempt:
Applying AM-GM:
$$f(x) \geq 2\sqrt{\operatorname{cosec}^2x\cdot25\cdot \sec^2x}$$
$$f(x) \geq 10\cdot \left|\operatorname{cosec}x\cdot \sec x\right| = \frac{20}{\left|\sin2x\right|}\geq20$$
Hence, my answer is $[20, \infty)$
But my book used the expansion for $\operatorname{cosec}^2x$ and $\sec^2x$ and arrived at $36$ as the minimum bound. I agree with their method as well, but I don't seem to find any problem my own method as well! 
Why am I getting a different result? What did I do wrong? Hints appreciated!
UPDATE: Thanks to explanation in comments, the problem appears to be that I've got a number for the range, but I haven't proved that it is the lowest number $f(x)$ can be equal to. It is entirely possible $f(x)$ can be greater than $30$ as well. We need to prove that as well. Now, I am stuck on this part, proving that $f(x)=20$ is possible or not. It seems the issue's like using two different values of $sin(2x)=1 and 5/13$ in the same question, the former as used above and the latter used for the fact that for AM-GM to actually take the minimum value, $\operatorname{cosec} ^2x=25\sec^2x$. Can there be more clarification on this?

REPLY [5 votes]: It's obvious that the maximum does not exist.
By C-S we obtain:
$$f(x)=(\sin^2x+\cos^2x)\left(\frac{1}{\sin^2x}+\frac{25}{\cos^2x}\right)\geq(1+5)^2=36.$$
The equality occurs for $(\sin{x},\cos{x})||\left(\frac{1}{\sin{x}},\frac{5}{\cos{x}}\right)$, which says that the equality occurs.
Thus, the answer is $[36,+\infty)$.