Mixing
Inequality properties between their multiplications [on hold]
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If $A>B>C>D$ then will $$(A/B * A/C * A/D) > (B/A * B/C * B/D) > (C/A * C/B * C/D) > (D/A * D/B * D/C)$$ be always true? If not, in what intervals will it not be true?
Obviously $A,B,C,D geq 0$ and they take values in the interval of ${Bbb R}_+$, that is positive real numbers only.
inequality universal-property
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Wuestenfux
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put on hold as unclear what you're asking by amWhy, Mark, Zvi, Rebellos, Shailesh 15 hours ago
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If $A>B>C>D$ then will $$(A/B * A/C * A/D) > (B/A * B/C * B/D) > (C/A * C/B * C/D) > (D/A * D/B * D/C)$$ be always true? If not, in what intervals will it not be true?
Obviously $A,B,C,D geq 0$ and they take values in the interval of ${Bbb R}_+$, that is positive real numbers only.
inequality universal-property
edited yesterday
Wuestenfux
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asked yesterday
mathamateur621
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put on hold as unclear what you're asking by amWhy, Mark, Zvi, Rebellos, Shailesh 15 hours ago
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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If $A>B>C>D$ then will $$(A/B * A/C * A/D) > (B/A * B/C * B/D) > (C/A * C/B * C/D) > (D/A * D/B * D/C)$$ be always true? If not, in what intervals will it not be true?
Obviously $A,B,C,D geq 0$ and they take values in the interval of ${Bbb R}_+$, that is positive real numbers only.
inequality universal-property
edited yesterday
Wuestenfux
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asked yesterday
mathamateur621
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mathamateur621 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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If $A>B>C>D$ then will $$(A/B * A/C * A/D) > (B/A * B/C * B/D) > (C/A * C/B * C/D) > (D/A * D/B * D/C)$$ be always true? If not, in what intervals will it not be true?
Obviously $A,B,C,D geq 0$ and they take values in the interval of ${Bbb R}_+$, that is positive real numbers only.
inequality universal-property
inequality universal-property
edited yesterday
Wuestenfux
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asked yesterday
mathamateur621
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mathamateur621 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited yesterday
Wuestenfux
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asked yesterday
mathamateur621
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edited yesterday
Wuestenfux
2,3541410
edited yesterday
Wuestenfux
2,3541410
edited yesterday
Wuestenfux
2,3541410
2,3541410
asked yesterday
mathamateur621
1
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asked yesterday
mathamateur621
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asked yesterday
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mathamateur621 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as unclear what you're asking by amWhy, Mark, Zvi, Rebellos, Shailesh 15 hours ago
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as unclear what you're asking by amWhy, Mark, Zvi, Rebellos, Shailesh 15 hours ago
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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First of all, we're dividing by $A,B,C,D$ so none can be $0$. Hence, in this answer, we assume $A,B,C,D>0$.
Anyway, since $A>B>C>D>0$, we know that
$$A^4>B^4>C^4>D^4$$
and now since $ABCD>0$, we may divide by $ABCD$ without flipping any signs:
$$frac{A^3}{BCD}>frac{B^3}{ACD}>frac{C^3}{ABD}>frac{D^3}{ABC}$$
which is the exact same as the thing you were trying to prove.
answered yesterday
vrugtehagel
10.7k1549
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Yes---compare term by term e.g. $A/B > B/A$, $A/C > B/C$, $A/D > B/D$ which will give you the first result, and so on.
answered yesterday
Richard Martin
1,3138
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
First of all, we're dividing by $A,B,C,D$ so none can be $0$. Hence, in this answer, we assume $A,B,C,D>0$.
Anyway, since $A>B>C>D>0$, we know that
$$A^4>B^4>C^4>D^4$$
and now since $ABCD>0$, we may divide by $ABCD$ without flipping any signs:
$$frac{A^3}{BCD}>frac{B^3}{ACD}>frac{C^3}{ABD}>frac{D^3}{ABC}$$
which is the exact same as the thing you were trying to prove.
answered yesterday
vrugtehagel
10.7k1549
add a comment |
up vote
1
down vote
First of all, we're dividing by $A,B,C,D$ so none can be $0$. Hence, in this answer, we assume $A,B,C,D>0$.
Anyway, since $A>B>C>D>0$, we know that
$$A^4>B^4>C^4>D^4$$
and now since $ABCD>0$, we may divide by $ABCD$ without flipping any signs:
$$frac{A^3}{BCD}>frac{B^3}{ACD}>frac{C^3}{ABD}>frac{D^3}{ABC}$$
which is the exact same as the thing you were trying to prove.
answered yesterday
vrugtehagel
10.7k1549
add a comment |
up vote
1
down vote
up vote
1
down vote
First of all, we're dividing by $A,B,C,D$ so none can be $0$. Hence, in this answer, we assume $A,B,C,D>0$.
Anyway, since $A>B>C>D>0$, we know that
$$A^4>B^4>C^4>D^4$$
and now since $ABCD>0$, we may divide by $ABCD$ without flipping any signs:
$$frac{A^3}{BCD}>frac{B^3}{ACD}>frac{C^3}{ABD}>frac{D^3}{ABC}$$
which is the exact same as the thing you were trying to prove.
answered yesterday
vrugtehagel
10.7k1549
First of all, we're dividing by $A,B,C,D$ so none can be $0$. Hence, in this answer, we assume $A,B,C,D>0$.
Anyway, since $A>B>C>D>0$, we know that
$$A^4>B^4>C^4>D^4$$
and now since $ABCD>0$, we may divide by $ABCD$ without flipping any signs:
$$frac{A^3}{BCD}>frac{B^3}{ACD}>frac{C^3}{ABD}>frac{D^3}{ABC}$$
which is the exact same as the thing you were trying to prove.
answered yesterday
vrugtehagel
10.7k1549
answered yesterday
vrugtehagel
10.7k1549
answered yesterday
vrugtehagel
10.7k1549
answered yesterday
vrugtehagel
10.7k1549
10.7k1549
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add a comment |
up vote
0
down vote
Yes---compare term by term e.g. $A/B > B/A$, $A/C > B/C$, $A/D > B/D$ which will give you the first result, and so on.
answered yesterday
Richard Martin
1,3138
add a comment |
up vote
0
down vote
Yes---compare term by term e.g. $A/B > B/A$, $A/C > B/C$, $A/D > B/D$ which will give you the first result, and so on.
answered yesterday
Richard Martin
1,3138
add a comment |
up vote
0
down vote
up vote
0
down vote
Yes---compare term by term e.g. $A/B > B/A$, $A/C > B/C$, $A/D > B/D$ which will give you the first result, and so on.
answered yesterday
Richard Martin
1,3138
Yes---compare term by term e.g. $A/B > B/A$, $A/C > B/C$, $A/D > B/D$ which will give you the first result, and so on.
answered yesterday
Richard Martin
1,3138
answered yesterday
Richard Martin
1,3138
answered yesterday
Richard Martin
1,3138
answered yesterday
Richard Martin
1,3138
1,3138
add a comment |
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