Prove this inequality with trigonometry $9\cos^2{x}-10\cos{x}\sin{y}-8\cos{y}\sin{x}+17\ge 1$ let$x,y\in R$,show that $$9\cos^2{x}-10\cos{x}\sin{y}-8\cos{y}\sin{x}+17\ge 1$$ Maybe use Cauchy-Schwarz inequality can solve it?and I can't Adit it:I think the right hand can replace constant $9$ ,and it's best

Let $u=9\cos^2x-10\cos x\sin y-8\cos y\sin x$

As $-\sqrt{a^2+b^2}\le-(a\cos v+b\sin v)\le\sqrt{a^2+b^2}$

$\implies-\sqrt{(10\cos x)^2+(8\sin x)^2}\le-(10\cos x\sin y+8\cos y\sin x)\le\sqrt{(10\cos x)^2+(8\sin x)^2}$

$u\ge9\cos^2x-\sqrt{(10\cos x)^2+(8\sin x)^2}=(\sqrt{9\cos^2x+16}-1)^2-17$

Now $3\le\sqrt{16}-1\le\sqrt{9\cos^2x+16}-1\le\sqrt{9+16}-1=4$

$\implies3^2\le(\sqrt{9\cos^2x+16}-1)^2\le4^2$

$\cdots$