Refinement of a strong inequality

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It's related to this If $a+b+c=abc$ then $sumlimits_{cyc}frac{1}{7a+b}leqfrac{sqrt3}{8}$ .
I make a little refinement wich could be usefull to prove the original one .

Let $a,b,c$ be real positive numbers such that $abc=a+b+c$ and $ageq b geq c$ then we have :
$$sum_{cyc}frac{1}{7a+b}leq frac{3}{8(frac{a+b+c}{3})}-3e^{frac{1}{8(frac{a+b+c}{3})}}+sum_{cyc}e^{frac{1}{7a+b}}leq frac{sqrt{3}}{{8}}$$

For the LHS it's just Jensen's inequality . To prove the RHS we need a strong inequality,stronger than Jensen's inequality I think .

Any hints would be appreciable.

Thanks

edited Nov 17 at 12:59

asked Nov 16 at 13:37

max8128

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add a comment |

up vote
-2
down vote

favorite

It's related to this If $a+b+c=abc$ then $sumlimits_{cyc}frac{1}{7a+b}leqfrac{sqrt3}{8}$ .
I make a little refinement wich could be usefull to prove the original one .

Let $a,b,c$ be real positive numbers such that $abc=a+b+c$ and $ageq b geq c$ then we have :
$$sum_{cyc}frac{1}{7a+b}leq frac{3}{8(frac{a+b+c}{3})}-3e^{frac{1}{8(frac{a+b+c}{3})}}+sum_{cyc}e^{frac{1}{7a+b}}leq frac{sqrt{3}}{{8}}$$

For the LHS it's just Jensen's inequality . To prove the RHS we need a strong inequality,stronger than Jensen's inequality I think .

Any hints would be appreciable.

Thanks

edited Nov 17 at 12:59

asked Nov 16 at 13:37

max8128

1,021421

add a comment |

up vote
-2
down vote

favorite

It's related to this If $a+b+c=abc$ then $sumlimits_{cyc}frac{1}{7a+b}leqfrac{sqrt3}{8}$ .
I make a little refinement wich could be usefull to prove the original one .

Let $a,b,c$ be real positive numbers such that $abc=a+b+c$ and $ageq b geq c$ then we have :
$$sum_{cyc}frac{1}{7a+b}leq frac{3}{8(frac{a+b+c}{3})}-3e^{frac{1}{8(frac{a+b+c}{3})}}+sum_{cyc}e^{frac{1}{7a+b}}leq frac{sqrt{3}}{{8}}$$

For the LHS it's just Jensen's inequality . To prove the RHS we need a strong inequality,stronger than Jensen's inequality I think .

Any hints would be appreciable.

Thanks

edited Nov 17 at 12:59

asked Nov 16 at 13:37

max8128

1,021421

It's related to this If $a+b+c=abc$ then $sumlimits_{cyc}frac{1}{7a+b}leqfrac{sqrt3}{8}$ .
I make a little refinement wich could be usefull to prove the original one .

Let $a,b,c$ be real positive numbers such that $abc=a+b+c$ and $ageq b geq c$ then we have :
$$sum_{cyc}frac{1}{7a+b}leq frac{3}{8(frac{a+b+c}{3})}-3e^{frac{1}{8(frac{a+b+c}{3})}}+sum_{cyc}e^{frac{1}{7a+b}}leq frac{sqrt{3}}{{8}}$$

For the LHS it's just Jensen's inequality . To prove the RHS we need a strong inequality,stronger than Jensen's inequality I think .

Any hints would be appreciable.

Thanks

real-analysis inequality contest-math jensen-inequality

edited Nov 17 at 12:59

asked Nov 16 at 13:37

max8128

1,021421

edited Nov 17 at 12:59

asked Nov 16 at 13:37

max8128

1,021421

edited Nov 17 at 12:59

asked Nov 16 at 13:37

max8128

1,021421

asked Nov 16 at 13:37

max8128

1,021421

asked Nov 16 at 13:37

max8128

1,021421

add a comment |

1 Answer
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Wolfram Alpha tells me that for $a=1$, $b=2$, $c=3$ you get ca. $0.2186$ in the middle and ca. $0.2165$ on the RHS, so this is false.

answered Nov 16 at 19:58

Michał Miśkiewicz

2,733616

Sorry I forget to add a condition...
– max8128
Nov 17 at 12:59

add a comment |

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1 Answer
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1 Answer
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active

oldest

votes

up vote
0
down vote

Wolfram Alpha tells me that for $a=1$, $b=2$, $c=3$ you get ca. $0.2186$ in the middle and ca. $0.2165$ on the RHS, so this is false.

answered Nov 16 at 19:58

Michał Miśkiewicz

2,733616

Sorry I forget to add a condition...
– max8128
Nov 17 at 12:59

add a comment |

up vote
0
down vote

Wolfram Alpha tells me that for $a=1$, $b=2$, $c=3$ you get ca. $0.2186$ in the middle and ca. $0.2165$ on the RHS, so this is false.

answered Nov 16 at 19:58

Michał Miśkiewicz

2,733616

Sorry I forget to add a condition...
– max8128
Nov 17 at 12:59

add a comment |

up vote
0
down vote

Wolfram Alpha tells me that for $a=1$, $b=2$, $c=3$ you get ca. $0.2186$ in the middle and ca. $0.2165$ on the RHS, so this is false.

answered Nov 16 at 19:58

Michał Miśkiewicz

2,733616

Wolfram Alpha tells me that for $a=1$, $b=2$, $c=3$ you get ca. $0.2186$ in the middle and ca. $0.2165$ on the RHS, so this is false.

answered Nov 16 at 19:58

Michał Miśkiewicz

2,733616

answered Nov 16 at 19:58

Michał Miśkiewicz

2,733616

answered Nov 16 at 19:58

Michał Miśkiewicz

2,733616

answered Nov 16 at 19:58

Michał Miśkiewicz

2,733616

Sorry I forget to add a condition...
– max8128
Nov 17 at 12:59

add a comment |

Sorry I forget to add a condition...
– max8128
Nov 17 at 12:59

Sorry I forget to add a condition...
– max8128
Nov 17 at 12:59

add a comment |

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