Does this inequality hold with some constant factor $c>0$? Does there exist a real number $c>0$ such that
$$ (x-1)^2+(y-1)^2-2(\sqrt{xy}-1)^2\ge c\big( (x-\sqrt{xy})^2+(y-\sqrt{xy})^2 \big) \tag{*}$$
holds for every positive real numbers $x,y$ such that $xy \ge \frac{1}{4}$.
Note that the LHS vanishes exactly when 
$$ (x-y)^2=2\big( (x+y)-2\sqrt{xy} \big),$$
which implies, since $xy \ge \frac{1}{4}$ that $x=y$ so the RHS vanishes as well.
Edit:
There seems to be some "symmetry imbalance" between the two sides of $(*)$. Indeed replacing $(x,y)$ by $ (\lambda x,\lambda y)$ multiplies the RHS by $\lambda^2$, but the LHS does not scale in this way exactly-some of its summands get multiplied by $\lambda$ and some $\lambda^2$. (after the $1$'s cancel each other).
Can this observation be lifted easily to a contradiction? 
 A: Substitute 
$$
r = 2\sqrt{xy} - 1
\qquad
q = x+y-2\sqrt{xy}.
$$
Thanks to AM-GM they are independent and can be any positive number.
It boils down to
$$
\frac{q+2r}{q+r+1}\ge c
$$
so $c$ cannot be positive, since for $q,r$ small enough, the expression converges to zero.

For example, if $x=1/2$ and $y=1/2+\varepsilon$, then you will find that $c(\varepsilon)$ goes to zero as $\varepsilon$ goes to zero.  

Here are the details. Expand
$$
(x-1)^2+(y-1)^2-2(\sqrt{xy}-1)^2\ge c\big( (x-\sqrt{xy})^2+(y-\sqrt{xy})^2 \big) $$
and obtain
$$
x^2 + y^2 -2x-2y-2xy+4\sqrt{xy}\ge c\big( x^2 + y^2 + 2xy -2x\sqrt{xy} -2y\sqrt{xy} \big).
$$
You can now regroup and factorize as follows
$$
\big( x+y-2\sqrt{xy}  \big)
(x+y+2\sqrt{xy} -2)
 \ge c(x+y)\big( x+y-2\sqrt{xy}  \big).
$$
If $x+y-2\sqrt{xy} = 0$, then the inequality is satisfied for every $c$, so we can semplify it. Thanks to AM-GM, $x+y-2\sqrt{xy} \ge 0$,so 
$$
x+y+2\sqrt{xy} -2
 \ge c(x+y).
$$
$x,y$ are positive, so we divide by $x+y$ and get
$$
\frac{x+y+2\sqrt{xy} -2}{x+y}\ge c
$$
and using the substitution
$$
r = 2\sqrt{xy} - 1
\qquad
q = x+y-2\sqrt{xy}
$$
we have
$$
\frac{q+2r}{q+r+1}\ge c.
$$
