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

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

$\cdots$

\begin{align*}9\cos^2x−10\cos x\sin y−8\cos y\sin x+17 &= 9\cos^2x−2\cos x\sin y−8(\cos x\sin y +\cos y\sin x)+17 \\ & = 9\cos^2x−2\cos x\sin y−8\sin(x+y)+17 \\ & \geq -2-8+17 \\ & \geq 7 \\ & \geq 1 \end{align*}

EDIT: I plugged the equation into wolfram alpha and told it to minimize it. Apparently the minimum is $9$. I'm fairly sure my work is correct now.
• $cos^2 x$ could be $0$. I ignored the $9$ since the inequality would hold anyways – Hwai-Ray Tung Jul 26 '16 at 4:35
• Nice,Now I have find the right is $9$ – communnites Jul 26 '16 at 4:42