Unsure if I have solved/proven this trigonometric inequality.

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