Minimizing $9 \sec^2{x} + 16 \csc^2{x}$

Question

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 ?

Answer 1

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.

Answer 2

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.

Answer 3

@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 $$9\sec^2 x + 16\csc^2 x \geq 25+ 24=49.$$