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.
 A: $f(t) = t\sin t,\  g(t) = \frac{t}{\sin t}$ 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 $xy$.
A: Here is another solution: let $f(t) = t + 1/t$, and observe that it is increasing on $[1,\infty)$, and decreasing on $(0,1]$.
Since $f(1/t) = f(t)$ for any $t > 0$, we can assume w.l.o.g. that $B \leq A$.
Now, observe that the function $g(t)=\frac{\sin(t)}{t}$ is decreasing on $(0,\frac{\pi}{2})$, and thus:
$$
g(A) \leq g(B) \implies 1\leq\frac{\sin(A)}{\sin(B)}\leq \frac{A}{B}.
$$
Finally
$$
f(\frac{\sin(A)}{\sin(B)})\leq f(\frac{A}{B}) \implies \frac{\sin(A)}{\sin(B)} + \frac{\sin(B)}{\sin(A)}\leq \frac{A}{B} + \frac{B}{A}.
$$
