<s>[INST] Does $(x-1)^2+(y-1)^2 \le c\big((x-y)^2+(xy-1)^2\big) $ hold? Does there exist a positive constant $c>0$ such that
$$(x-1)^2+(y-1)^2 \le c\big((x-y)^2+(xy-1)^2\big) \tag{1}$$
holds for any nonnegative $x,y$?
Let me add some context for this question:
The motivation comes from the case were $x,y$ are interpreted as singular values of a $2 \times 2$ matrix $A$ with nonnegative determinant. Then $f(x,y)=(x-1)^2+(y-1)^2=\operatorname{dist}^2(A,\operatorname{SO}(2))$.
I am interested in bounding $\operatorname{dist}^2(A,\operatorname{SO}(2))$ from above by a sum of two terms: a term which penalizes deviations of $A$ from being area-preserving, and a term $\operatorname{dist}^2(A,\operatorname{CO}(2))$, which penalizes deviations from being conformal. (Here $\operatorname{CO}(2)=R^{+}\operatorname{SO}(2)$ is the group of conformal matrices).
In an answer to this previous question of mine, the following bound was proved:
$$
 (x-1)^2+(y-1)^2 \le |x-y||x+y| + 2|xy-1|.
$$
While this is close to what I had in mind, the term $|x-y||x+y|$ can be large even when $x,y$ become very close. In fact, one can prove that $\operatorname{dist}^2(A,\operatorname{CO}(2))=\frac{1}{2}(\sigma_1(A)-\sigma_2(A))^2$, so this is the reason for asking about the specific bound $(1)$. (The term $(x-y)^2$ corresponds to $\operatorname{dist}^2(A,\operatorname{CO}(2))$).
 [/INST] Let $x=y=0,$ we get $c \geqslant 2.$
For $c =2,$ inequality become
$$(x-1)^2+(y-1)^2 \leqslant 2\big((x-y)^2+(xy-1)^2\big),$$
equivalent to
$$2x^2y^2+x^2+y^2+2(x+y) \geqslant 8xy.$$
Using the AM-GM inequality, we have
$$2x^2y^2+x^2+y^2+2(x+y) \geqslant 8\sqrt[8]{(x^2y^2)^2 \cdot x^2 \cdot y^2 \cdot (xy)^2}=8xy.$$
So, your inequality holds for all $c \geqslant 2.$
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