Unsure if I have solved/proven this trigonometric inequality.

This is my first post here. I apologize if it goes against any guidelines for posting. I study math as a hobby and am currently dealing with trigonometry on a high school level. I have so far learned the formulas for trigonometric addition and subtraction and double the angle, as well as what is in my language referred to as the ’trigonometric one’ - getting the radius of the unit circle by use of the pythagorean theorem. I have not yet gotten to deriving trigonometric functions. The following is a problem I could solve by plugging in a set of numbers, but in seeking a more elegant solution, perhaps, I found myself stuck and I don’t know what I am missing. I am appreciative of any help I get. The problem is as follows:

Show that if $A$ is an angle and $0^\circ<A<90^\circ$ then $\hspace{0.3cm}\left( 1+\dfrac {1}{\sin A}\right) \left( 1+\dfrac {1}{\cos A}\right)>5$

I began with the following assumption:

$$0^\circ<A<90^\circ\rightarrow0<{\sin A}<1\\0<{\cos A}<1\rightarrow\dfrac {1}{\sin A}\\\dfrac {1}{\cos A}>1$$

Given the above, it would follow that:

$$\begin{aligned} \lim _{A\rightarrow 90^\circ}\dfrac {1}{\cos A}&=\infty \\ \lim _{A\rightarrow 0^\circ}\dfrac {1}{\sin A}&=\infty \end{aligned}$$

This alone doesn’t seem like enough to show what is asked. I can show that at $A=45^\circ$ the product is still greater than 5, but I am not sure how any offset in degrees from there affects two trigonometric terms such that the product is still greater than 5. I also tried solving the inequality but ended up with fractioned terms I couldn’t add up or a cubic function if you will, that I couldn’t solve.

They probably expected something like this:

Expanding out, your inequality is the same as $$1 + {1 \over \sin A} + {1 \over \cos A} + {1 \over \sin A \cos A} \geq 5$$ This is equivalent to $${1 \over \sin A} + {1 \over \cos A} + {1 \over \sin A \cos A} \geq 4$$ Which is the same as $${1 \over \sin A} + {1 \over \cos A} + {2 \over \sin 2A} \geq 4$$ Since in the range in question, $0 < \sin A, \cos A < 1$ and $0 < \sin 2A \leq 1$, one has $${1 \over \sin A} + {1 \over \cos A} + {2 \over \sin 2A} > 1 + 1 + 2$$ $$= 4$$

If we open the parenthesis, we get $$1+{1\over \sin A}+ {1\over \cos A}+{1\over \sin A \cos A}\geq 5$$ $${1+\sin A+ \cos A\over \sin A \cos A}\geq 4$$ $$1+\sin A+ \cos A -4 \sin A \cos A \ge 0$$ $$(\cos A - \sin A)^2+\cos A(1-\sin A)+\sin A(1-\cos A) \ge 0$$ In the last inequality, each term is not negative so the sum is not negative