Prove this inequality $\sum\limits_\text{cyc}\sqrt{1-xy}\ge 2$ Let $x,y,z\ge 0$, and $x+y+z=2$, show that
$$\sqrt{1-xy}+\sqrt{1-yz}+\sqrt{1-xz}\ge 2.$$
Mt try: $$\Longleftrightarrow 3-(xy+yz+zx)+2\sum_\text{cyc}\sqrt{(1-xy)(1-yz)}\ge 4$$
or
$$\sum_\text{cyc}\sqrt{(1-xy)(1-yz)}\ge\dfrac{1}{2}(1+xy+yz+zx)$$

WLOG, assume that $x \ge y \ge z$.
We have
$$\sqrt{1 - xy} \ge \sqrt{1 - (x + y)^2/4}.$$
Also, we have
$$\sqrt{1 - yz} + \sqrt{1 - zx} \ge 1 + \sqrt{1 - yz - zx}.$$
(Note: Squaring both sides, it is just
$2\sqrt{1 - yz - zx + z^2xy} \ge 2\sqrt{1 - yz - zx}$.)
Thus, it suffices to prove that
$$\sqrt{1 - (x + y)^2/4} + 1 + \sqrt{1 - yz - zx} \ge 2$$
or
$$\sqrt{1 - (2 - z)^2/4} + \sqrt{1 - (2 - z)z} \ge 1$$
or
$$\frac12\sqrt{z(4 - z)} + (1 - z) \ge 1$$
which is true (using $z \le 2/3$).
We are done.
