# Prove that function $f(x)=\frac{27}{\sin(x)}+\frac{64}{\cos(x)}$ has a minimum value but no maximum value at $0<x<\frac{\pi}{2}$.

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 $$64\sin^3(x)-27\cos^3(x)<0,\: \text{if}\ 0 $$64\sin^3(x)-27\cos^3(x)>0,\: \text{if}\ 0 and I got stuck.

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