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
$$\sum_\text{cyc}\sqrt{(1-xy)(1-yz)}\ge\dfrac{1}{2}(1+xy+yz+zx)$$
 A: After squaring of the both sides we need to prove that
$$2\sum\limits_{cyc}\sqrt{(1-xy)(1-xz)}\geq1+xy+xz+yz$$ or
$$2\sum\limits_{cyc}\sqrt{((x+y+z)^2-4xy)((x+y+z)^2-4xz)}\geq(x+y+z)^2+4(xy+xz+yz)$$ or
$$2\sum\limits_{cyc}\sqrt{((x-y)^2+z^2+2xz+2yz)((x-z)^2+y^2+2xy+2yz)}\geq\sum\limits_{cyc}(x^2+6xy)$$
and after using C-S it remains to prove that
$$2\sum_{cyc}((x-y)(x-z)+xy+2x\sqrt{yz}+2xy)\geq\sum\limits_{cyc}(x^2+6xy)$$ or
$$\sum_{cyc}(x^2-2xy+4x\sqrt{yz})\geq0$$ or
$$\sum\limits_{cyc}\left(x^2-\sqrt{x^3y}-\sqrt{x^3z}+x\sqrt{yz}\right)+\sum_{cyc}\left(\sqrt{x^3y}+\sqrt{x^3z}-2xy\right)+3\sum_{cyc}x\sqrt{yz}\geq0$$
which is Schur and Muirhead.
A: Lets minimise the LHS.  WLOG let $x \ge y \ge z \ge 0$ - i.e. $x \in [y, 2], y \in [z, x], z \in [0, y]$.  
The LHS is concave in all variables, hence its minima will happen when $x, y, z $ take extreme values in their allowable intervals.  Along with the constraint, this leads to only three possibilities viz. $(x, y, z) \in \{(2, 0, 0), (1, 1, 0), (\frac23, \frac23, \frac23)\}$
In particular note that the minimum is given when $x=y=1, z=0$ (or permutations of it).
A: Lets assume $z=\min\{x,y,z\}$ then note $\sqrt{1-xy}+\sqrt{1-yz}+\sqrt{1-xz}\geq\sqrt{1-\frac{(2-z)^2}{4}}+2\sqrt[4]{(1-yz)(1-xz)}=\sqrt{z-\frac{z^2}{4}}+2\sqrt[4]{1-z(2-z)+z^2xy}\geq\sqrt{z-\frac{z^2}{4}}+2\sqrt[4]{(1-z)^2+z^4}\geq 2$
A: 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.
