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 ?

@Michael Rozenberg has already given an answer using Cauchy-Schwarz inequality. Here's my answer using the AM-GM inequality.
Write $\sec^2 x = 1 + \tan^2 x$ and $\csc^2 x = 1 + \cot^2 x$. Then
$$9\sec^2 x + 16\csc^2 x = 9(1+\tan^2 x) + 16(1+ \cot^2 x) = 25 + 9\tan^2 x + \frac{16}{\tan^2 x}.$$
Then by AM-GM inequality
$$9\tan^2 x + \frac{16}{\tan ^2 x} \geq 2\sqrt{9\tan^2 x\cdot\frac{16}{\tan^2 x}}$$
$$=2\sqrt{9\cdot 16}=2\cdot3\cdot4 = 24.$$
Thus 
$$9\sec^2 x + 16\csc^2 x \geq 25+ 24=49.$$
