Minimizing $9 \sec^2{x} + 16 \csc^2{x}$ Find the minimum value of $$9 \sec^2{x} + 16 \csc^2{x}$$
My turn :
Using AM-GM
$$9\sec^2{x} + 16\csc^2{x} \geq 2 \sqrt{144 \sec^2{x} \csc^2{x}}$$
$$9 \sec^2{x} + 16\csc^2{x} \geq 24 \sec{x} \csc{x} $$
But the equality sign holds iff
$$9 \sec^2{x} = 16\csc^2{x}$$
Then $$ \tan{x} = \frac{4}{3}$$
Then the minimum value is $$24 \times \frac{5}{4} \times \frac{5}{3} = 50$$
Is there any mistake with the solution ?

Your mistake in the following.
You proved that $$\frac{9}{\cos^2x}+\frac{16}{\sin^2x}\geq\frac{24}{|\sin{x}\cos{x}|}=\frac{48}{|\sin2x|},$$
but you did not find a minimal value.
After your first step we see that $$\frac{48}{|\sin2x|}\geq48,$$ but it does not give a minimal value because the value $48$ does not occur.
One of right solutions is the following.
By C-S
$$\frac{9}{\cos^2x}+\frac{16}{\sin^2x}=(\cos^2x+\sin^2x)\left(\frac{9}{\cos^2x}+\frac{16}{\sin^2x}\right)\geq(3+4)^2=49.$$
The equality occurs for $(\cos{x},\sin{x})||\left(\frac{3}{\cos{x}},\frac{4}{\sin{x}}\right),$ which says that we got a minimal value.
