Prove the inequality $sum_{cyc} {{a+abc} over {1+ab+abcd}} ge {{10} over {3}}$ with Cauchy-Schwarz [closed]
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Problem: If $abcde = 1$, $a, b, c, d, e > 0$, $a, b, c, d, e in Bbb R$, prove that
$sum_{cyc} {{a+abc} over {1+ab+abcd}} ge {{10} over {3}}$
First I proceeded with Cauchy-Schwartz inequality, but I couldn't find any way to use that. I think that the equal condition is $a=b=c=d=e=1$.
I don't know what to do to solve this problem. Can somebody please give me a hint?
inequality substitution cauchy-schwarz-inequality
edited Jan 11 at 7:24
Michael Rozenberg
99k1590189
asked Jan 11 at 6:49
coding1101coding1101
394
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closed as off-topic by Carl Mummert, A. Pongrácz, Lord_Farin, José Carlos Santos, clathratus Jan 15 at 0:58
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Carl Mummert, A. Pongrácz, Lord_Farin, José Carlos Santos, clathratus
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Problem: If $abcde = 1$, $a, b, c, d, e > 0$, $a, b, c, d, e in Bbb R$, prove that
$sum_{cyc} {{a+abc} over {1+ab+abcd}} ge {{10} over {3}}$
First I proceeded with Cauchy-Schwartz inequality, but I couldn't find any way to use that. I think that the equal condition is $a=b=c=d=e=1$.
I don't know what to do to solve this problem. Can somebody please give me a hint?
inequality substitution cauchy-schwarz-inequality
edited Jan 11 at 7:24
Michael Rozenberg
99k1590189
asked Jan 11 at 6:49
coding1101coding1101
394
$endgroup$
closed as off-topic by Carl Mummert, A. Pongrácz, Lord_Farin, José Carlos Santos, clathratus Jan 15 at 0:58
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Carl Mummert, A. Pongrácz, Lord_Farin, José Carlos Santos, clathratus
If this question can be reworded to fit the rules in the help center, please edit the question.
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Okay. Thank you for noticing me.
$endgroup$
– coding1101
Jan 11 at 6:53
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What does $sum_{cyc}$ mean?
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– Chickenmancer
Jan 11 at 7:09
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It means that we add it by a cycle. The problem is equivalent to ${{a+abc} over {1+ab+abcd}}+{{b+bcd} over {1+bc+bcde}}+{{c+cde} over {1+cd+cdea}}+{{d+dea} over {1+de+deab}}+{{e+eab} over {1+ea+eabc}} ge {10 over 3}$.
$endgroup$
– coding1101
Jan 11 at 7:13
add a comment |
$begingroup$
Problem: If $abcde = 1$, $a, b, c, d, e > 0$, $a, b, c, d, e in Bbb R$, prove that
$sum_{cyc} {{a+abc} over {1+ab+abcd}} ge {{10} over {3}}$
First I proceeded with Cauchy-Schwartz inequality, but I couldn't find any way to use that. I think that the equal condition is $a=b=c=d=e=1$.
I don't know what to do to solve this problem. Can somebody please give me a hint?
inequality substitution cauchy-schwarz-inequality
edited Jan 11 at 7:24
Michael Rozenberg
99k1590189
asked Jan 11 at 6:49
coding1101coding1101
394
$endgroup$
Problem: If $abcde = 1$, $a, b, c, d, e > 0$, $a, b, c, d, e in Bbb R$, prove that
$sum_{cyc} {{a+abc} over {1+ab+abcd}} ge {{10} over {3}}$
First I proceeded with Cauchy-Schwartz inequality, but I couldn't find any way to use that. I think that the equal condition is $a=b=c=d=e=1$.
I don't know what to do to solve this problem. Can somebody please give me a hint?
inequality substitution cauchy-schwarz-inequality
inequality substitution cauchy-schwarz-inequality
edited Jan 11 at 7:24
Michael Rozenberg
99k1590189
asked Jan 11 at 6:49
coding1101coding1101
394
edited Jan 11 at 7:24
Michael Rozenberg
99k1590189
asked Jan 11 at 6:49
coding1101coding1101
394
edited Jan 11 at 7:24
Michael Rozenberg
99k1590189
edited Jan 11 at 7:24
Michael Rozenberg
99k1590189
edited Jan 11 at 7:24
Michael Rozenberg
99k1590189
99k1590189
asked Jan 11 at 6:49
coding1101coding1101
394
asked Jan 11 at 6:49
coding1101coding1101
394
asked Jan 11 at 6:49
coding1101coding1101
394
394
closed as off-topic by Carl Mummert, A. Pongrácz, Lord_Farin, José Carlos Santos, clathratus Jan 15 at 0:58
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Carl Mummert, A. Pongrácz, Lord_Farin, José Carlos Santos, clathratus
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Carl Mummert, A. Pongrácz, Lord_Farin, José Carlos Santos, clathratus Jan 15 at 0:58
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Carl Mummert, A. Pongrácz, Lord_Farin, José Carlos Santos, clathratus
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Okay. Thank you for noticing me.
$endgroup$
– coding1101
Jan 11 at 6:53
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What does $sum_{cyc}$ mean?
$endgroup$
– Chickenmancer
Jan 11 at 7:09
$begingroup$
It means that we add it by a cycle. The problem is equivalent to ${{a+abc} over {1+ab+abcd}}+{{b+bcd} over {1+bc+bcde}}+{{c+cde} over {1+cd+cdea}}+{{d+dea} over {1+de+deab}}+{{e+eab} over {1+ea+eabc}} ge {10 over 3}$.
$endgroup$
– coding1101
Jan 11 at 7:13
add a comment |
$begingroup$
Okay. Thank you for noticing me.
$endgroup$
– coding1101
Jan 11 at 6:53
$begingroup$
What does $sum_{cyc}$ mean?
$endgroup$
– Chickenmancer
Jan 11 at 7:09
$begingroup$
It means that we add it by a cycle. The problem is equivalent to ${{a+abc} over {1+ab+abcd}}+{{b+bcd} over {1+bc+bcde}}+{{c+cde} over {1+cd+cdea}}+{{d+dea} over {1+de+deab}}+{{e+eab} over {1+ea+eabc}} ge {10 over 3}$.
$endgroup$
– coding1101
Jan 11 at 7:13
$begingroup$
Okay. Thank you for noticing me.
$endgroup$
– coding1101
Jan 11 at 6:53
$begingroup$
Okay. Thank you for noticing me.
$endgroup$
– coding1101
Jan 11 at 6:53
$begingroup$
What does $sum_{cyc}$ mean?
$endgroup$
– Chickenmancer
Jan 11 at 7:09
$begingroup$
What does $sum_{cyc}$ mean?
$endgroup$
– Chickenmancer
Jan 11 at 7:09
$begingroup$
It means that we add it by a cycle. The problem is equivalent to ${{a+abc} over {1+ab+abcd}}+{{b+bcd} over {1+bc+bcde}}+{{c+cde} over {1+cd+cdea}}+{{d+dea} over {1+de+deab}}+{{e+eab} over {1+ea+eabc}} ge {10 over 3}$.
$endgroup$
– coding1101
Jan 11 at 7:13
$begingroup$
It means that we add it by a cycle. The problem is equivalent to ${{a+abc} over {1+ab+abcd}}+{{b+bcd} over {1+bc+bcde}}+{{c+cde} over {1+cd+cdea}}+{{d+dea} over {1+de+deab}}+{{e+eab} over {1+ea+eabc}} ge {10 over 3}$.
$endgroup$
– coding1101
Jan 11 at 7:13
add a comment |
1 Answer
1
active
oldest
votes
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Yes, C-S helps!
Indeed, let $a=frac{y}{z},$ $b=frac{z}{y}$, $c=frac{t}{z}$, $d=frac{w}{t},$ where $x$, $y$, $z$, $t$ and $w$ are positives.
Thus, the condition gives $e=frac{x}{w}$ and we obtain:
$$sum_{cyc}frac{a+abc}{1+ab+abcd}=sum_{cyc}frac{frac{y}{x}+frac{y}{x}cdotfrac{z}{y}cdotfrac{t}{z}}{1+frac{y}{x}cdotfrac{z}{y}+frac{y}{x}cdotfrac{z}{y}cdotfrac{t}{z}cdotfrac{w}{t}}=$$
$$=sum_{cyc}frac{frac{y}{x}+frac{t}{x}}{1+frac{z}{x}+frac{w}{x}}=sum_{cyc}frac{y+t}{x+z+w}=-5+sum_{cyc}frac{x+y+z+t+w}{x+z+w}=$$
$$=-5+frac{1}{3}sum_{cyc}(x+z+w)sum_{cyc}frac{1}{x+z+w}.$$
Can you end it now?
answered Jan 11 at 7:24
Michael RozenbergMichael Rozenberg
99k1590189
$endgroup$
$begingroup$
Thank you for helping me! By the way, isn't $a = frac{y}{x}$? And I think the fourth equation is $sum_{cyc} frac{y+t}{x+z+w}$.
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– coding1101
Jan 11 at 7:41
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@coding1101 It was typo. Thank you! I fixed. You are welcome!
$endgroup$
– Michael Rozenberg
Jan 11 at 7:44
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes, C-S helps!
Indeed, let $a=frac{y}{z},$ $b=frac{z}{y}$, $c=frac{t}{z}$, $d=frac{w}{t},$ where $x$, $y$, $z$, $t$ and $w$ are positives.
Thus, the condition gives $e=frac{x}{w}$ and we obtain:
$$sum_{cyc}frac{a+abc}{1+ab+abcd}=sum_{cyc}frac{frac{y}{x}+frac{y}{x}cdotfrac{z}{y}cdotfrac{t}{z}}{1+frac{y}{x}cdotfrac{z}{y}+frac{y}{x}cdotfrac{z}{y}cdotfrac{t}{z}cdotfrac{w}{t}}=$$
$$=sum_{cyc}frac{frac{y}{x}+frac{t}{x}}{1+frac{z}{x}+frac{w}{x}}=sum_{cyc}frac{y+t}{x+z+w}=-5+sum_{cyc}frac{x+y+z+t+w}{x+z+w}=$$
$$=-5+frac{1}{3}sum_{cyc}(x+z+w)sum_{cyc}frac{1}{x+z+w}.$$
Can you end it now?
answered Jan 11 at 7:24
Michael RozenbergMichael Rozenberg
99k1590189
$endgroup$
$begingroup$
Thank you for helping me! By the way, isn't $a = frac{y}{x}$? And I think the fourth equation is $sum_{cyc} frac{y+t}{x+z+w}$.
$endgroup$
– coding1101
Jan 11 at 7:41
$begingroup$
@coding1101 It was typo. Thank you! I fixed. You are welcome!
$endgroup$
– Michael Rozenberg
Jan 11 at 7:44
add a comment |
$begingroup$
Yes, C-S helps!
Indeed, let $a=frac{y}{z},$ $b=frac{z}{y}$, $c=frac{t}{z}$, $d=frac{w}{t},$ where $x$, $y$, $z$, $t$ and $w$ are positives.
Thus, the condition gives $e=frac{x}{w}$ and we obtain:
$$sum_{cyc}frac{a+abc}{1+ab+abcd}=sum_{cyc}frac{frac{y}{x}+frac{y}{x}cdotfrac{z}{y}cdotfrac{t}{z}}{1+frac{y}{x}cdotfrac{z}{y}+frac{y}{x}cdotfrac{z}{y}cdotfrac{t}{z}cdotfrac{w}{t}}=$$
$$=sum_{cyc}frac{frac{y}{x}+frac{t}{x}}{1+frac{z}{x}+frac{w}{x}}=sum_{cyc}frac{y+t}{x+z+w}=-5+sum_{cyc}frac{x+y+z+t+w}{x+z+w}=$$
$$=-5+frac{1}{3}sum_{cyc}(x+z+w)sum_{cyc}frac{1}{x+z+w}.$$
Can you end it now?
answered Jan 11 at 7:24
Michael RozenbergMichael Rozenberg
99k1590189
$endgroup$
$begingroup$
Thank you for helping me! By the way, isn't $a = frac{y}{x}$? And I think the fourth equation is $sum_{cyc} frac{y+t}{x+z+w}$.
$endgroup$
– coding1101
Jan 11 at 7:41
$begingroup$
@coding1101 It was typo. Thank you! I fixed. You are welcome!
$endgroup$
– Michael Rozenberg
Jan 11 at 7:44
add a comment |
$begingroup$
Yes, C-S helps!
Indeed, let $a=frac{y}{z},$ $b=frac{z}{y}$, $c=frac{t}{z}$, $d=frac{w}{t},$ where $x$, $y$, $z$, $t$ and $w$ are positives.
Thus, the condition gives $e=frac{x}{w}$ and we obtain:
$$sum_{cyc}frac{a+abc}{1+ab+abcd}=sum_{cyc}frac{frac{y}{x}+frac{y}{x}cdotfrac{z}{y}cdotfrac{t}{z}}{1+frac{y}{x}cdotfrac{z}{y}+frac{y}{x}cdotfrac{z}{y}cdotfrac{t}{z}cdotfrac{w}{t}}=$$
$$=sum_{cyc}frac{frac{y}{x}+frac{t}{x}}{1+frac{z}{x}+frac{w}{x}}=sum_{cyc}frac{y+t}{x+z+w}=-5+sum_{cyc}frac{x+y+z+t+w}{x+z+w}=$$
$$=-5+frac{1}{3}sum_{cyc}(x+z+w)sum_{cyc}frac{1}{x+z+w}.$$
Can you end it now?
answered Jan 11 at 7:24
Michael RozenbergMichael Rozenberg
99k1590189
$endgroup$
Yes, C-S helps!
Indeed, let $a=frac{y}{z},$ $b=frac{z}{y}$, $c=frac{t}{z}$, $d=frac{w}{t},$ where $x$, $y$, $z$, $t$ and $w$ are positives.
Thus, the condition gives $e=frac{x}{w}$ and we obtain:
$$sum_{cyc}frac{a+abc}{1+ab+abcd}=sum_{cyc}frac{frac{y}{x}+frac{y}{x}cdotfrac{z}{y}cdotfrac{t}{z}}{1+frac{y}{x}cdotfrac{z}{y}+frac{y}{x}cdotfrac{z}{y}cdotfrac{t}{z}cdotfrac{w}{t}}=$$
$$=sum_{cyc}frac{frac{y}{x}+frac{t}{x}}{1+frac{z}{x}+frac{w}{x}}=sum_{cyc}frac{y+t}{x+z+w}=-5+sum_{cyc}frac{x+y+z+t+w}{x+z+w}=$$
$$=-5+frac{1}{3}sum_{cyc}(x+z+w)sum_{cyc}frac{1}{x+z+w}.$$
Can you end it now?
answered Jan 11 at 7:24
Michael RozenbergMichael Rozenberg
99k1590189
edited Jan 11 at 7:42
answered Jan 11 at 7:24
Michael RozenbergMichael Rozenberg
99k1590189
answered Jan 11 at 7:24
Michael RozenbergMichael Rozenberg
99k1590189
answered Jan 11 at 7:24
Michael RozenbergMichael Rozenberg
99k1590189
99k1590189
$begingroup$
Thank you for helping me! By the way, isn't $a = frac{y}{x}$? And I think the fourth equation is $sum_{cyc} frac{y+t}{x+z+w}$.
$endgroup$
– coding1101
Jan 11 at 7:41
$begingroup$
@coding1101 It was typo. Thank you! I fixed. You are welcome!
$endgroup$
– Michael Rozenberg
Jan 11 at 7:44
add a comment |
$begingroup$
Thank you for helping me! By the way, isn't $a = frac{y}{x}$? And I think the fourth equation is $sum_{cyc} frac{y+t}{x+z+w}$.
$endgroup$
– coding1101
Jan 11 at 7:41
$begingroup$
@coding1101 It was typo. Thank you! I fixed. You are welcome!
$endgroup$
– Michael Rozenberg
Jan 11 at 7:44
$begingroup$
Thank you for helping me! By the way, isn't $a = frac{y}{x}$? And I think the fourth equation is $sum_{cyc} frac{y+t}{x+z+w}$.
$endgroup$
– coding1101
Jan 11 at 7:41
$begingroup$
Thank you for helping me! By the way, isn't $a = frac{y}{x}$? And I think the fourth equation is $sum_{cyc} frac{y+t}{x+z+w}$.
$endgroup$
– coding1101
Jan 11 at 7:41
$begingroup$
@coding1101 It was typo. Thank you! I fixed. You are welcome!
$endgroup$
– Michael Rozenberg
Jan 11 at 7:44
$begingroup$
@coding1101 It was typo. Thank you! I fixed. You are welcome!
$endgroup$
– Michael Rozenberg
Jan 11 at 7:44
add a comment |
$begingroup$
Okay. Thank you for noticing me.
$endgroup$
– coding1101
Jan 11 at 6:53
$begingroup$
What does $sum_{cyc}$ mean?
$endgroup$
– Chickenmancer
Jan 11 at 7:09
$begingroup$
It means that we add it by a cycle. The problem is equivalent to ${{a+abc} over {1+ab+abcd}}+{{b+bcd} over {1+bc+bcde}}+{{c+cde} over {1+cd+cdea}}+{{d+dea} over {1+de+deab}}+{{e+eab} over {1+ea+eabc}} ge {10 over 3}$.
$endgroup$
– coding1101
Jan 11 at 7:13