Find the minimum of the value $f=(1+\sin^2{x})(1+\sin^2{y})$

Question

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?

Answer 1

Let $\tan^2x=a$ and $\tan^2y=b$.

Thus, $ab=\frac{1}{4}$ and by AM-GM we obtain: $$(1+\sin^2x)(1+\sin^2y)=(2-\cos^2x)(2-\cos^2y)=$$ $$=\left(2-\frac{1}{1+a}\right)\left(2-\frac{1}{1+b}\right)=\frac{(2a+1)(2b+1)}{(1+a)(1+b)}=$$ $$=\frac{2+2(a+b)}{\frac{5}{4}+a+b}=2-\frac{\frac{1}{2}}{\frac{5}{4}+a+b}\geq$$ $$\geq2-\frac{\frac{1}{2}}{\frac{5}{4}+2\sqrt{ab}}=\frac{16}{9}.$$ The equality occurs for example for $\tan{x}=\frac{1}{\sqrt2}$ and $\tan{y}=-\frac{1}{\sqrt2}$,

which says that we got a minimal value.

Answer 2

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 $f>0, 9f-16\ge0\iff f\ge?$