<s>[INST] 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 ?
 [/INST] By differentiating and setting to $0$ or plotting the graph, the minimum value is actually $49$, so there must be something wrong in your solution.
From $9 \sec^2 x = 16 \csc^2 x$ you cannot deduce $\tan x = \frac{4}{3}$. You have $\tan^2 x = \frac{16}{9}$, but $\tan x$ can either be $\frac{4}{3}$ or $-\frac{4}{3}$.
In addition, $24 \sec x \csc x$ is actually negative for some values of $x$, such as $x = 2$. Therefore you have not actually proven that $50$ is the minimum value, but rather that the function is always greater than some negative number.
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