Inequality with roots and bivariate quadratic forms
For $|a|le 1$ and $|b|le 1$ show that
$$sqrt{1 - b^2}sqrt{1 - a} le frac{sqrt{6}(1-frac{3}{4}b)(7a^2b^2 + 96a^2b + 65a^2 + 14ab^2 + 42a + 7b^2 - 96b - 135)}{43a^2b^2 + 129a^2 + 86ab^2 - 86a + 43b^2 - 327}
$$
Numerical checks confirm this involved inequality. I tried expansions of the roots and clearing denominators, but didn't succeed. It holds with equality for $a=b=1$ and $a=-b=1$. Hints are welcome. Since numerator and denominator are bivariate quadratic forms, also a geometric interpretation might be helpful.
inequality quadratic-forms
asked yesterday
Andreas
7,8281037
add a comment |
For $|a|le 1$ and $|b|le 1$ show that
$$sqrt{1 - b^2}sqrt{1 - a} le frac{sqrt{6}(1-frac{3}{4}b)(7a^2b^2 + 96a^2b + 65a^2 + 14ab^2 + 42a + 7b^2 - 96b - 135)}{43a^2b^2 + 129a^2 + 86ab^2 - 86a + 43b^2 - 327}
$$
Numerical checks confirm this involved inequality. I tried expansions of the roots and clearing denominators, but didn't succeed. It holds with equality for $a=b=1$ and $a=-b=1$. Hints are welcome. Since numerator and denominator are bivariate quadratic forms, also a geometric interpretation might be helpful.
inequality quadratic-forms
asked yesterday
Andreas
7,8281037
quadratic form? Degree varies from 4 to zero.
– Macavity
yesterday
add a comment |
For $|a|le 1$ and $|b|le 1$ show that
$$sqrt{1 - b^2}sqrt{1 - a} le frac{sqrt{6}(1-frac{3}{4}b)(7a^2b^2 + 96a^2b + 65a^2 + 14ab^2 + 42a + 7b^2 - 96b - 135)}{43a^2b^2 + 129a^2 + 86ab^2 - 86a + 43b^2 - 327}
$$
Numerical checks confirm this involved inequality. I tried expansions of the roots and clearing denominators, but didn't succeed. It holds with equality for $a=b=1$ and $a=-b=1$. Hints are welcome. Since numerator and denominator are bivariate quadratic forms, also a geometric interpretation might be helpful.
inequality quadratic-forms
asked yesterday
Andreas
7,8281037
For $|a|le 1$ and $|b|le 1$ show that
$$sqrt{1 - b^2}sqrt{1 - a} le frac{sqrt{6}(1-frac{3}{4}b)(7a^2b^2 + 96a^2b + 65a^2 + 14ab^2 + 42a + 7b^2 - 96b - 135)}{43a^2b^2 + 129a^2 + 86ab^2 - 86a + 43b^2 - 327}
$$
Numerical checks confirm this involved inequality. I tried expansions of the roots and clearing denominators, but didn't succeed. It holds with equality for $a=b=1$ and $a=-b=1$. Hints are welcome. Since numerator and denominator are bivariate quadratic forms, also a geometric interpretation might be helpful.
inequality quadratic-forms
inequality quadratic-forms
asked yesterday
Andreas
7,8281037
asked yesterday
Andreas
7,8281037
edited yesterday
asked yesterday
Andreas
7,8281037
asked yesterday
Andreas
7,8281037
asked yesterday
Andreas
7,8281037
7,8281037
quadratic form? Degree varies from 4 to zero.
– Macavity
yesterday
add a comment |
quadratic form? Degree varies from 4 to zero.
– Macavity
yesterday
quadratic form? Degree varies from 4 to zero.
– Macavity
yesterday
quadratic form? Degree varies from 4 to zero.
– Macavity
yesterday
add a comment |
1 Answer
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no idea. The denominator really is nonzero on your square, namely less than zero
answered yesterday
Will Jagy
102k599199
Thanks. The numerator is also nonpositive.
– Andreas
yesterday
I was firstly looking for a general way to approach such a problem, the full solution could wait .....
– Andreas
yesterday
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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no idea. The denominator really is nonzero on your square, namely less than zero
answered yesterday
Will Jagy
102k599199
Thanks. The numerator is also nonpositive.
– Andreas
yesterday
I was firstly looking for a general way to approach such a problem, the full solution could wait .....
– Andreas
yesterday
add a comment |
no idea. The denominator really is nonzero on your square, namely less than zero
answered yesterday
Will Jagy
102k599199
Thanks. The numerator is also nonpositive.
– Andreas
yesterday
I was firstly looking for a general way to approach such a problem, the full solution could wait .....
– Andreas
yesterday
add a comment |
no idea. The denominator really is nonzero on your square, namely less than zero
answered yesterday
Will Jagy
102k599199
no idea. The denominator really is nonzero on your square, namely less than zero
answered yesterday
Will Jagy
102k599199
answered yesterday
Will Jagy
102k599199
answered yesterday
Will Jagy
102k599199
answered yesterday
Will Jagy
102k599199
102k599199
Thanks. The numerator is also nonpositive.
– Andreas
yesterday
I was firstly looking for a general way to approach such a problem, the full solution could wait .....
– Andreas
yesterday
add a comment |
Thanks. The numerator is also nonpositive.
– Andreas
yesterday
I was firstly looking for a general way to approach such a problem, the full solution could wait .....
– Andreas
yesterday
Thanks. The numerator is also nonpositive.
– Andreas
yesterday
Thanks. The numerator is also nonpositive.
– Andreas
yesterday
I was firstly looking for a general way to approach such a problem, the full solution could wait .....
– Andreas
yesterday
I was firstly looking for a general way to approach such a problem, the full solution could wait .....
– Andreas
yesterday
add a comment |
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quadratic form? Degree varies from 4 to zero.
– Macavity
yesterday