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By differentiation I got the derivative $$f'(x)=\frac{64\sin^3(x)-27\cos^3(x)}{\sin^2(x)\cos^2(x)}$$ and then got the zero of derivative $$x=\arctan(\frac{3}{4})$$ insert x to f(x)=y get the "minimum" value $$f_{min}(x)=125$$ but I don't know how to prove this value is exactly the minimum not the maximum, noticed that the denominator of $f'(x)$ always be positive, then I only need to prove $$64\sin^3(x)-27\cos^3(x)<0,\: \text{if}\ 0<x<\arctan(3/4)$$ $$64\sin^3(x)-27\cos^3(x)>0,\: \text{if}\ 0<x<\arctan(3/4)$$ and I got stuck.

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2 Answers 2

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Never mind :), guess this is the benefit of rewriting, forces one to think deeper and may inspire oneself to dig out the answer. By the way, if any one get a better answer, please post it, I'll read it, Thanks!


case one: $x<arctan(3/4)$ $$ \begin{align} 64sin^3(x)-27cos^3(x)&<0\\ tan(x)&<\frac{3}{4}\\ \end{align} $$ this is obviously true because $tan(x)$ is monotonic increasing on $(0,\frac{\pi}{2})$

case two is similar to case one.

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The maximum value of the expression is $\infty$ when $x=0$ or any other angle in periodicity. This means that either of the denominators has to be $0$. Since when $$x\in\left(0,\frac{\pi}2\right)$$ the denominator i.e. $\operatorname{sin}x$ or $\operatorname{cos}x$ can never be $0$ therefore the answer that you got by differentiating $(125)$ is the minimum.

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