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If $a+b+c=abc$ then $sumlimits_cycfrac17a+bleqfracsqrt38$

11

7

$begingroup$

Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=abc$. Prove that:
$$frac17a+b+frac17b+c+frac17c+aleqfracsqrt38$$

I tried C-S:
$$left(sum_cycfrac17a+bright)^2leqsum_cycfrac1(ka+mb+c)(7a+b)^2sum_cyc(ka+mb+c)=$$
$$=sum_cycfrac(k+m+1)(a+b+c)(ka+mb+c)(7a+b)^2.$$
Thus, it remains to prove that
$$sum_cycfrack+m+1(ka+mb+c)(7a+b)^2leqfrac364abc,$$
but I did not find non-negative values of $k$ and $m$, for which the last inequality is true.

If we replace $7$ with $8$ so for $(a,b,c)||(28,1,5)$ this inequality would be wrong. Around this point the starting inequality is true, but we see that we can'not free use AM-GM because in AM-GM the equality occurs, when all variables are equal.

Thank you!

inequality contest-math

share|cite|improve this question

edited Mar 25 at 11:30

Michael Rozenberg

asked Mar 21 '17 at 20:05

Michael RozenbergMichael Rozenberg

110k1896201

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  FOr those fans who hate abbreviations, C-S is Cauchy-Schwartz
  $endgroup$
  – Mark Fischler
  Mar 21 '17 at 20:10
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  As a good start, take $a=sqrt3x, b = sqrt3y, c = sqrt3z$ so that you are working with $x+y+z = 3xyz$ and the extremum will occur at $(1,1,1)$. Makes the horrendous algebra a little cleaner.
  $endgroup$
  – Mark Fischler
  Mar 21 '17 at 22:57
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @W-t-P It's cyclic and not symmetric.
  $endgroup$
  – Michael Rozenberg
  Mar 24 at 20:42
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MichaelRozenberg Maybe BW helps? Though it doesn't seem to be particularly elegant...
  $endgroup$
  – user574848
  Apr 1 at 10:18
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @user574848 I tried. I think BW does not help here.
  $endgroup$
  – Michael Rozenberg
  Apr 1 at 20:57
  

  
  
  
  

 | 
show 2 more comments

11

7

$begingroup$

Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=abc$. Prove that:
$$frac17a+b+frac17b+c+frac17c+aleqfracsqrt38$$

I tried C-S:
$$left(sum_cycfrac17a+bright)^2leqsum_cycfrac1(ka+mb+c)(7a+b)^2sum_cyc(ka+mb+c)=$$
$$=sum_cycfrac(k+m+1)(a+b+c)(ka+mb+c)(7a+b)^2.$$
Thus, it remains to prove that
$$sum_cycfrack+m+1(ka+mb+c)(7a+b)^2leqfrac364abc,$$
but I did not find non-negative values of $k$ and $m$, for which the last inequality is true.

If we replace $7$ with $8$ so for $(a,b,c)||(28,1,5)$ this inequality would be wrong. Around this point the starting inequality is true, but we see that we can'not free use AM-GM because in AM-GM the equality occurs, when all variables are equal.

Thank you!

inequality contest-math

share|cite|improve this question

edited Mar 25 at 11:30

Michael Rozenberg

asked Mar 21 '17 at 20:05

Michael RozenbergMichael Rozenberg

110k1896201

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  FOr those fans who hate abbreviations, C-S is Cauchy-Schwartz
  $endgroup$
  – Mark Fischler
  Mar 21 '17 at 20:10
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  As a good start, take $a=sqrt3x, b = sqrt3y, c = sqrt3z$ so that you are working with $x+y+z = 3xyz$ and the extremum will occur at $(1,1,1)$. Makes the horrendous algebra a little cleaner.
  $endgroup$
  – Mark Fischler
  Mar 21 '17 at 22:57
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @W-t-P It's cyclic and not symmetric.
  $endgroup$
  – Michael Rozenberg
  Mar 24 at 20:42
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MichaelRozenberg Maybe BW helps? Though it doesn't seem to be particularly elegant...
  $endgroup$
  – user574848
  Apr 1 at 10:18
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @user574848 I tried. I think BW does not help here.
  $endgroup$
  – Michael Rozenberg
  Apr 1 at 20:57
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=abc$. Prove that:
$$frac17a+b+frac17b+c+frac17c+aleqfracsqrt38$$

I tried C-S:
$$left(sum_cycfrac17a+bright)^2leqsum_cycfrac1(ka+mb+c)(7a+b)^2sum_cyc(ka+mb+c)=$$
$$=sum_cycfrac(k+m+1)(a+b+c)(ka+mb+c)(7a+b)^2.$$
Thus, it remains to prove that
$$sum_cycfrack+m+1(ka+mb+c)(7a+b)^2leqfrac364abc,$$
but I did not find non-negative values of $k$ and $m$, for which the last inequality is true.

If we replace $7$ with $8$ so for $(a,b,c)||(28,1,5)$ this inequality would be wrong. Around this point the starting inequality is true, but we see that we can'not free use AM-GM because in AM-GM the equality occurs, when all variables are equal.

Thank you!

inequality contest-math

share|cite|improve this question

edited Mar 25 at 11:30

Michael Rozenberg

asked Mar 21 '17 at 20:05

Michael RozenbergMichael Rozenberg

110k1896201

$endgroup$

share|cite|improve this question

edited Mar 25 at 11:30

Michael Rozenberg

asked Mar 21 '17 at 20:05

Michael RozenbergMichael Rozenberg

110k1896201

share|cite|improve this question

edited Mar 25 at 11:30

Michael Rozenberg

asked Mar 21 '17 at 20:05

Michael RozenbergMichael Rozenberg

110k1896201

asked Mar 21 '17 at 20:05

Michael RozenbergMichael Rozenberg

110k1896201

asked Mar 21 '17 at 20:05

Michael RozenbergMichael Rozenberg

110k1896201