How prove this $\sqrt{a^2(\cos{x}+\cos{y})^2+b^2(\sin{x}+\sin{y})^2}+\sqrt{a^2(\cos{x}-\cos{y})^2+b^2(\sin{x}-\sin{y})^2}\le 2\sqrt{a^2+b^2}$ let $a,b>0$ is give numbers,for any $x,y\in R$,show that
$$\sqrt{a^2(\cos{x}+\cos{y})^2+b^2(\sin{x}+\sin{y})^2}+\sqrt{a^2(\cos{x}-\cos{y})^2+b^2(\sin{x}-\sin{y})^2}\le 2\sqrt{a^2+b^2}$$
maybe can use Cauchy-Schwarz inequality to solve it.
This inequality background is from:How prove this $|ON|\le \sqrt{a^2+b^2}$
$$\Longleftrightarrow 2a^2(\cos^2{x}+\cos^2{y})+2b^2(\sin^2{x}+\sin^2{y})+2\sqrt{a^2(\cos{x}+\cos{y})^2+b^2(\sin{x}+\sin{y})^2}\sqrt{a^2(\cos{x}-\cos{y})^2+b^2(\sin{x}-\sin{y})^2}\le 4(a^2+b^2)$$
But I can't,Thank you for you help
 A: Let us make use of the trigonometric identities for the sum of angles i.e.
$\cos(a \pm b) = \cos a \cos b \mp \sin a \sin b$
$\sin(a \pm b) = \sin a \cos b \pm \cos a \sin b$
Thus 
$\cos x+ \cos y = 2\cos (\frac{x+y}{2})\cos (\frac{x-y}{2})$
$\cos x- \cos y = -2\sin (\frac{x+y}{2})\sin (\frac{x-y}{2})$
$\sin x+ \sin y = 2\sin (\frac{x+y}{2})\cos (\frac{x-y}{2})$
$\sin x- \sin y = 2\cos (\frac{x+y}{2})\sin (\frac{x-y}{2})$
Going back to the LHS of the inequality, we have 
$2\cos (\frac{x-y}{2})\sqrt{a^2\cos^2 (\frac{x+y}{2})+b^2\sin^2 (\frac{x+y}{2})}+2\sin (\frac{x-y}{2})\sqrt{a^2\sin^2 (\frac{x+y}{2})+b^2\cos^2 (\frac{x+y}{2})}$
We are in a position to use the Cauchy-Schwarz inequality, which we state below for reference
$\sum_{k=1}^nf_kg_k\leq (\sum_{k=1}^nf_k^2)^{1/2}(\sum_{k=1}^ng_k^2)^{1/2}$
If we let $f_1=2\cos (\frac{x-y}{2})$ and $f_2=2\sin (\frac{x-y}{2})$
and we let 
$g_1=\sqrt{a^2\cos^2 (\frac{x+y}{2})+b^2\sin^2 (\frac{x+y}{2})}$ and
$g_2=\sqrt{a^2\sin^2 (\frac{x+y}{2})+b^2\cos^2 (\frac{x+y}{2})}$
The upper bound of the inequality is
$\sqrt{4\cos^2 (\frac{x-y}{2})+4\sin^2 (\frac{x-y}{2})}\sqrt{a^2\cos^2 (\frac{x+y}{2})+b^2\sin^2 (\frac{x+y}{2})+a^2\sin^2 (\frac{x+y}{2})+b^2\cos^2 (\frac{x+y}{2})}$
and recalling that $\sin^2 \theta + \cos^2 \theta = 1$, the upper bound becomes
$2\sqrt{a^2+b^2}$
which is what we set out to prove.
