Let $0<x,y<\frac {\pi}{2}$ such that $\sin (x+y)=\frac 23$, then find the minimum of
$$\frac{\sin x}{\cos y}+\frac {\cos x}{\sin y}+\frac{\sin y}{\cos x}+\frac{\cos y}{\sin x}$$
A) $\frac 23$
B) $\frac 43$
C) $\frac 89$
D) $\frac {16}{9}$
E) $\frac{32}{27}$
My attempts:
I think that the all possible answers are wrong. Because, by Am-Gm inequality we have
$$\frac{\sin x}{\cos y}+\frac{\cos y}{\sin x}+\frac {\cos x}{\sin y}+\frac{\sin y}{\cos x}≥2+2=4.$$
But, Wolfram Alpha gives us a different result : The Global Minimum doesn't exist. However, the local minimum must be $6.$
But, still the problem is not solved. Because Wolfram's graph shows that the minimum can be less than $6$.
I also tried
Let $$\sin x=a,\cos y=b,\cos x=c,\sin y=d$$ with
$$ab+cd=\frac 23≥2\sqrt{abcd}\implies abcd≤\frac 19\\ a^2+c^2=b^2+d^2=1 $$
then I need
$$\min \left(\frac ab+\frac ba+\frac cd+\frac dc\right)$$
But, I can't do anything from here. Finally, I attach the graph drawn by WA.
Best Answer
Simplifying gives: $$\frac{\sin x}{\cos y}+\frac {\cos x}{\sin y}+\frac{\sin y}{\cos x}+\frac{\cos y}{\sin x}=\frac{\sin^2 x+\cos^2y}{\sin x\cos y}+\frac {\cos^2 x+\sin^2y}{\sin y\cos x}=\frac{\sin(x+y)\sin(x-y)+1}{\sin x\cos y}+\frac {1-\sin(x+y)\sin(x-y)}{\sin y\cos x}=\frac {8}{3\sin2x\sin 2y}+\frac{2\sin (x-y)}{3}\left(\frac{1}{\sin x\cos y}-\frac{1}{\sin y\cos x}\right)=\frac {8}{3\sin2x\sin 2y}-\frac 83\frac{\sin^2(x-y)}{\sin2x\sin 2y}=\frac{8\cos^2(x-y)}{3\sin 2x\sin2y}=\frac{16\cos^2(x-y)}{3(\cos 2(x-y)-\cos 2(x+y))}=\frac{16}{3(2-10/9\sec^2(x-y))}$$
The denominator is maximum when $\sec^2(x-y)$ is minimum, which happens when $x=y$, in which case the quantity has value $6$. So the required global minimum should be $6$.
Edit: The edit shows that the given expression (let's call it $f(x,y)$ for brevity) can't take any value smaller than $6$. To see that, let's first note (Refer Note) that $\color{blue}{ 2-\frac {10}9\sec^2(x-y)\gt 0 \text{ for all } (x,y)\in (0,\frac \pi 2)\times (0,\frac \pi 2)}$.
Suppose on the contrary that there exist some $a$ and $b$ in $(0,\frac \pi 2)$ such that $f(a,b)<6$. It follows that $$\frac{16}{3(2-10/9\sec^2(x-y))}\lt 6\implies 8<18-10\sec^2(x-y)\implies \sec^2(x-y)\lt 1 $$ and this is a contradiction. So the assumption that $f$ takes any value less than $6$ is wrong. Hence global minimum value of $f$ under the given conditions is $6$.
Note: If $\sec^2(x-y)\ge \frac{18}{10}$ then $\cos^2(x-y)\le \frac {10}{18}\implies 0\lt \cos (x-y)\le \sqrt {\frac{10}{18}}$ (because $x-y \in (-\frac \pi 2, \frac \pi 2)$ so $\cos $ is +ve) so let's consider two cases: 1) $0\le x-y)\lt \frac \pi 2 \text{ and } 2) -\frac \pi 2\lt x-y \lt 0$.
Case 1: it follows that $x-y\ge \arccos\sqrt {\frac{10}{18}}$ (noting that $\cos$ is decreasing on $[0,\pi/2))$. Given that $x+y=\arcsin \frac 23$, it follows that $y\le 0$ which is not possible.
Case 2: it follows that $x-y\le -\arccos\sqrt {\frac{10}{18}}$ (noting that $\cos$ is increasing on $(-\frac \pi 2, 0))$. Then proceeding as in case 1) results in a contradiction.
This establishes the blue colored part.