How prove this inequality $\sum\limits_{cyc}\sqrt{\frac{\cos{A}\cos{B}}{\cos{C}}}\ge\frac{3\sqrt{2}}{2}$ let $x,y,z>0$,and such $$x^2y^2+y^2z^2+x^2z^2+2x^2y^2z^2=1$$
show that
$$x+y+z\ge\dfrac{3\sqrt{2}}{2}$$
My idea: in $\Delta ABC$,we have 
$$\cos^2{A}+\cos^2{B}+\cos^2{C}+2\cos{A}\cos{B}\cos{C}=1$$
so let
$$xy=\cos{A},yz=\cos{B},zx=\cos{C},A,B,C\in(0,\dfrac{\pi}{2})$$
$$\Longleftrightarrow \sqrt{\dfrac{\cos{B}\cos{C}}{\cos{A}}}+\sqrt{\dfrac{\cos{A}\cos{C}}{\cos{B}}}+\sqrt{\dfrac{\cos{A}\cos{B}}{\cos{C}}}\ge\dfrac{3\sqrt{2}}{2}$$
But I can't.Thank you  very much

The claim is false: If we let $x = 0$, $y=1$ and $z=1$, then we have
$x+y+z = 2 < 3 \sqrt{2} / 2$. (This conclusion holds by continuity for some $x,y,z > 0$, too.)
