<s>[INST] Prove that $\frac{\sin A}{\sin B}+\frac{\sin B}{\sin A} \leq \frac{A}{B}+\frac{B}{A}$ for acute angles, $A$ and $B$. 
Prove that $\frac{\sin A}{\sin B}+\frac{\sin B}{\sin A} \leq \frac{A}{B}+\frac{B}{A}$ for acute angles, $A$ and $B$.

I'm confused about how to do this since we can't say $\frac{\sin A}{\sin B}\leq \frac{A}{B}$. So I simplified and got  $$\frac{\sin^2 A+ \sin^2 B}{\sin A \sin B} \leq \frac{A^2+B^2}{AB}$$
Using $\sin x \leq x$ we can say $\sin^2 A+ \sin^2 B \le A^2+B^2$ but since we cannot divide, this doesn't work either.
 [/INST] $f(t) = t\sin t,\  g(t) = \frac{t}{\sin t}$ are both increasing functions on $(0,\pi/2)$.
Also note that: if $a \le b$ and $c\le d$, $$ad +bc \le ac+bd\tag{1}$$ which follows from $(a-b)(c-d) \ge 0$.
Suppose $x <y$. Then we have $x\sin x <y \sin y$ and $\frac{x}{\sin x} < \frac{y}{\sin y}$.
Then from $(1)$, it follows that
$$(x \sin x)\frac{y}{\sin y} +(y \sin y)\frac{x}{\sin x} \le (x \sin x)\frac{x}{\sin x} + (y \sin y)\frac{y}{\sin y} = x^2+y^2$$
Your inequality follows after dividing both sides by $xy$.
</s>