Inequality with roots and bivariate quadratic forms
$begingroup$
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 Jan 5 at 0:44
AndreasAndreas
8,4161137
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add a comment |
$begingroup$
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 Jan 5 at 0:44
AndreasAndreas
8,4161137
$endgroup$
$begingroup$
quadratic form? Degree varies from 4 to zero.
$endgroup$
– Macavity
Jan 5 at 18:27
$begingroup$
@Macavity Yes. I phrased that somewhat loosely "bivariate quadratic forms" to indicate that the two variables occur at most quadratic. I was trying to avoid "quartic form" since then a single variable can occur in fourth power. If there is a correct term I'm happy to change the formulation.
$endgroup$
– Andreas
Jan 7 at 8:44
add a comment |
$begingroup$
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 Jan 5 at 0:44
AndreasAndreas
8,4161137
$endgroup$
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 Jan 5 at 0:44
AndreasAndreas
8,4161137
asked Jan 5 at 0:44
AndreasAndreas
8,4161137
edited Jan 5 at 0:51
Andreas
asked Jan 5 at 0:44
AndreasAndreas
8,4161137
asked Jan 5 at 0:44
AndreasAndreas
8,4161137
asked Jan 5 at 0:44
AndreasAndreas
8,4161137
8,4161137
$begingroup$
quadratic form? Degree varies from 4 to zero.
$endgroup$
– Macavity
Jan 5 at 18:27
$begingroup$
@Macavity Yes. I phrased that somewhat loosely "bivariate quadratic forms" to indicate that the two variables occur at most quadratic. I was trying to avoid "quartic form" since then a single variable can occur in fourth power. If there is a correct term I'm happy to change the formulation.
$endgroup$
– Andreas
Jan 7 at 8:44
add a comment |
$begingroup$
quadratic form? Degree varies from 4 to zero.
$endgroup$
– Macavity
Jan 5 at 18:27
$begingroup$
@Macavity Yes. I phrased that somewhat loosely "bivariate quadratic forms" to indicate that the two variables occur at most quadratic. I was trying to avoid "quartic form" since then a single variable can occur in fourth power. If there is a correct term I'm happy to change the formulation.
$endgroup$
– Andreas
Jan 7 at 8:44
$begingroup$
quadratic form? Degree varies from 4 to zero.
$endgroup$
– Macavity
Jan 5 at 18:27
$begingroup$
quadratic form? Degree varies from 4 to zero.
$endgroup$
– Macavity
Jan 5 at 18:27
$begingroup$
@Macavity Yes. I phrased that somewhat loosely "bivariate quadratic forms" to indicate that the two variables occur at most quadratic. I was trying to avoid "quartic form" since then a single variable can occur in fourth power. If there is a correct term I'm happy to change the formulation.
$endgroup$
– Andreas
Jan 7 at 8:44
$begingroup$
@Macavity Yes. I phrased that somewhat loosely "bivariate quadratic forms" to indicate that the two variables occur at most quadratic. I was trying to avoid "quartic form" since then a single variable can occur in fourth power. If there is a correct term I'm happy to change the formulation.
$endgroup$
– Andreas
Jan 7 at 8:44
add a comment |
1 Answer
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no idea. The denominator really is nonzero on your square, namely less than zero
answered Jan 5 at 1:21
Will JagyWill Jagy
104k5102201
$endgroup$
$begingroup$
Thanks. The numerator is also nonpositive.
$endgroup$
– Andreas
Jan 5 at 15:11
$begingroup$
I was firstly looking for a general way to approach such a problem, the full solution could wait .....
$endgroup$
– Andreas
Jan 5 at 15:12
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
no idea. The denominator really is nonzero on your square, namely less than zero
answered Jan 5 at 1:21
Will JagyWill Jagy
104k5102201
$endgroup$
$begingroup$
Thanks. The numerator is also nonpositive.
$endgroup$
– Andreas
Jan 5 at 15:11
$begingroup$
I was firstly looking for a general way to approach such a problem, the full solution could wait .....
$endgroup$
– Andreas
Jan 5 at 15:12
add a comment |
$begingroup$
no idea. The denominator really is nonzero on your square, namely less than zero
answered Jan 5 at 1:21
Will JagyWill Jagy
104k5102201
$endgroup$
$begingroup$
Thanks. The numerator is also nonpositive.
$endgroup$
– Andreas
Jan 5 at 15:11
$begingroup$
I was firstly looking for a general way to approach such a problem, the full solution could wait .....
$endgroup$
– Andreas
Jan 5 at 15:12
add a comment |
$begingroup$
no idea. The denominator really is nonzero on your square, namely less than zero
answered Jan 5 at 1:21
Will JagyWill Jagy
104k5102201
$endgroup$
no idea. The denominator really is nonzero on your square, namely less than zero
answered Jan 5 at 1:21
Will JagyWill Jagy
104k5102201
answered Jan 5 at 1:21
Will JagyWill Jagy
104k5102201
answered Jan 5 at 1:21
Will JagyWill Jagy
104k5102201
answered Jan 5 at 1:21
Will JagyWill Jagy
104k5102201
104k5102201
$begingroup$
Thanks. The numerator is also nonpositive.
$endgroup$
– Andreas
Jan 5 at 15:11
$begingroup$
I was firstly looking for a general way to approach such a problem, the full solution could wait .....
$endgroup$
– Andreas
Jan 5 at 15:12
add a comment |
$begingroup$
Thanks. The numerator is also nonpositive.
$endgroup$
– Andreas
Jan 5 at 15:11
$begingroup$
I was firstly looking for a general way to approach such a problem, the full solution could wait .....
$endgroup$
– Andreas
Jan 5 at 15:12
$begingroup$
Thanks. The numerator is also nonpositive.
$endgroup$
– Andreas
Jan 5 at 15:11
$begingroup$
Thanks. The numerator is also nonpositive.
$endgroup$
– Andreas
Jan 5 at 15:11
$begingroup$
I was firstly looking for a general way to approach such a problem, the full solution could wait .....
$endgroup$
– Andreas
Jan 5 at 15:12
$begingroup$
I was firstly looking for a general way to approach such a problem, the full solution could wait .....
$endgroup$
– Andreas
Jan 5 at 15:12
add a comment |
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$begingroup$
quadratic form? Degree varies from 4 to zero.
$endgroup$
– Macavity
Jan 5 at 18:27
$begingroup$
@Macavity Yes. I phrased that somewhat loosely "bivariate quadratic forms" to indicate that the two variables occur at most quadratic. I was trying to avoid "quartic form" since then a single variable can occur in fourth power. If there is a correct term I'm happy to change the formulation.
$endgroup$
– Andreas
Jan 7 at 8:44