Find the minimum of the value $f=(1+\sin^2{x})(1+\sin^2{y})$ Let $\tan{x}\tan{y}=-\frac{1}{2}$,find the minumin of the value $$f=(1+\sin^2{x})(1+\sin^2{y})$$
My ugly solution:
$$f=\dfrac{1+\sin^2{x}}{\sin^2{x}+\cos^2{x}}\cdot\dfrac{1+\sin^2{y}}{\sin^2{y}+\cos^2{y}}=\dfrac{2m^2+1}{m^2+1}\cdot\dfrac{2n^2+1}{n^2+1}$$where $m=\tan{x},n=\tan{y},mn=-\frac{1}{2}$,so 
$$f=2\cdot\dfrac{4m^4+4m^2+1}{4m^4+5m^2+1}=2\left(1-\dfrac{m^2}{4m^4+1+5m^2}\right)\ge 2\left(1-\dfrac{m^2}{4m^2+5m^2}\right)=\dfrac{16}{9}$$
some other simple methods?such as AM-GM and Cauchy-Schwarz inequality kill it?

You have done the hard work.
On rearrangement
$$m^4(8-4f)+m^2(8-5f)+2-f=0$$ which is a Quadratic Equation in $m^2$ which is real
So, the discriminant must be $\ge0$
$$(8-5f)^2-4(8-4f)(2-f)=-16f+9f^2\ge0$$
$\iff f(9f-16)\ge0$
As $f>0, 9f-16\ge0\iff f\ge?$
