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Why is a

Absolute inequality derivationReducing a complex expression to a real expressionProve inequality statementsProve the following Inequality.Inequality on Quadratic Equation CoefficientsQuadratic mean inequality where arguments are linearly combination with 1Solve $sqrt{n^3+1}+sqrt{n^3}>10^3$Proof of simple inequality using only fundamental axiomsProving the following quadratic inequality?Prove that this expression is real and positive

3 Answers
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$begingroup$

If $a$ is positive, the left side is a positive number (pos/pos = pos) and the right side is a negative number (neg/pos = neg), so the inequality can never be satisfied (pos < neg is never true).

You can also eliminate the possibility that $a=0$ since neither side is then defined (division by zero would occur on both sides, which is not permitted).

The remaining possibility is that $a$ is negative.

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answered Mar 19 at 15:14

MPWMPW

31k12157

$endgroup$

add a comment | 

1

$begingroup$

One fraction is just the negative of the other. So (assuming $b^2 - a>0$) it becomes a matter of figuring out which is positive and which is negative. Since square roots by definition are positive, we get ...

share|cite|improve this answer

answered Mar 19 at 15:08

ArthurArthur

122k7122211

$endgroup$

add a comment | 

0

$begingroup$

First, note that for real values of $b$ that $b^2 geq 0$ thus any $a < 0$ will make the $b^2-a$ term a positive number. The rest follows by considering the denominator and the sign change in the numerator.

share|cite|improve this answer

answered Mar 19 at 15:05

CyclotomicFieldCyclotomicField

2,4331314

$endgroup$

add a comment | 

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0

$begingroup$

If $a$ is positive, the left side is a positive number (pos/pos = pos) and the right side is a negative number (neg/pos = neg), so the inequality can never be satisfied (pos < neg is never true).

You can also eliminate the possibility that $a=0$ since neither side is then defined (division by zero would occur on both sides, which is not permitted).

The remaining possibility is that $a$ is negative.

share|cite|improve this answer

answered Mar 19 at 15:14

MPWMPW

31k12157

$endgroup$

add a comment | 

0

$begingroup$

If $a$ is positive, the left side is a positive number (pos/pos = pos) and the right side is a negative number (neg/pos = neg), so the inequality can never be satisfied (pos < neg is never true).

You can also eliminate the possibility that $a=0$ since neither side is then defined (division by zero would occur on both sides, which is not permitted).

The remaining possibility is that $a$ is negative.

share|cite|improve this answer

answered Mar 19 at 15:14

MPWMPW

31k12157

$endgroup$

add a comment | 

$begingroup$

If $a$ is positive, the left side is a positive number (pos/pos = pos) and the right side is a negative number (neg/pos = neg), so the inequality can never be satisfied (pos < neg is never true).

You can also eliminate the possibility that $a=0$ since neither side is then defined (division by zero would occur on both sides, which is not permitted).

The remaining possibility is that $a$ is negative.

share|cite|improve this answer

answered Mar 19 at 15:14

MPWMPW

31k12157

$endgroup$

share|cite|improve this answer

answered Mar 19 at 15:14

MPWMPW

31k12157

answered Mar 19 at 15:14

MPWMPW

31k12157

answered Mar 19 at 15:14

MPWMPW

31k12157

1

$begingroup$

One fraction is just the negative of the other. So (assuming $b^2 - a>0$) it becomes a matter of figuring out which is positive and which is negative. Since square roots by definition are positive, we get ...

share|cite|improve this answer

answered Mar 19 at 15:08

ArthurArthur

122k7122211

$endgroup$

add a comment | 

1

$begingroup$

One fraction is just the negative of the other. So (assuming $b^2 - a>0$) it becomes a matter of figuring out which is positive and which is negative. Since square roots by definition are positive, we get ...

share|cite|improve this answer

answered Mar 19 at 15:08

ArthurArthur

122k7122211

$endgroup$

add a comment | 

$begingroup$

One fraction is just the negative of the other. So (assuming $b^2 - a>0$) it becomes a matter of figuring out which is positive and which is negative. Since square roots by definition are positive, we get ...

share|cite|improve this answer

answered Mar 19 at 15:08

ArthurArthur

122k7122211

$endgroup$

share|cite|improve this answer

answered Mar 19 at 15:08

ArthurArthur

122k7122211

answered Mar 19 at 15:08

ArthurArthur

122k7122211

answered Mar 19 at 15:08

ArthurArthur

122k7122211

0

$begingroup$

First, note that for real values of $b$ that $b^2 geq 0$ thus any $a < 0$ will make the $b^2-a$ term a positive number. The rest follows by considering the denominator and the sign change in the numerator.

share|cite|improve this answer

answered Mar 19 at 15:05

CyclotomicFieldCyclotomicField

2,4331314

$endgroup$

add a comment | 

0

$begingroup$

First, note that for real values of $b$ that $b^2 geq 0$ thus any $a < 0$ will make the $b^2-a$ term a positive number. The rest follows by considering the denominator and the sign change in the numerator.

share|cite|improve this answer

answered Mar 19 at 15:05

CyclotomicFieldCyclotomicField

2,4331314

$endgroup$

add a comment | 

$begingroup$

First, note that for real values of $b$ that $b^2 geq 0$ thus any $a < 0$ will make the $b^2-a$ term a positive number. The rest follows by considering the denominator and the sign change in the numerator.

share|cite|improve this answer

answered Mar 19 at 15:05

CyclotomicFieldCyclotomicField

2,4331314

$endgroup$

share|cite|improve this answer

answered Mar 19 at 15:05

CyclotomicFieldCyclotomicField

2,4331314

answered Mar 19 at 15:05

CyclotomicFieldCyclotomicField

2,4331314

answered Mar 19 at 15:05

CyclotomicFieldCyclotomicField

2,4331314