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