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?
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 $$ \frac{q+2r}{q+r+1}\ge c. $$