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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 $\operatorname{dist}^2(A,\operatorname{CO}(2))$).

Asaf Shachar
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  • You have asked many good questions earlier but this one is out of context. – Kavi Rama Murthy Aug 20 '20 at 07:34
  • Thanks for your comment. You are right. I appreciate that you commented here and not just downvoted (if you did so). Anyway, I agree that "being a good contributor" for the site and community should not give you immunity and free ability to post bad questions. To be honest, I wasn't sure that the specific context would interest too many people. But this might be just an excuse for laziness on my part. Anyway, I have now added some context. I am still not sure this question is really "of broad enough interest" to the community, but I guess we shall see:). Thanks again for contacting me. – Asaf Shachar Aug 20 '20 at 08:13
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    I did not vote for closure. I saw two votes for closure and thought I should help you by suggesting that you provide some context (so that the question does not get closed). – Kavi Rama Murthy Aug 20 '20 at 08:16

1 Answers1

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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 $c \geqslant 2.$

nguyenhuyenag
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