If $a+b+c=abc$ then $sumlimits_{cyc}frac{1}{7a+b}leqfrac{sqrt3}{8}$
$begingroup$
Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=abc$. Prove that:
$$frac{1}{7a+b}+frac{1}{7b+c}+frac{1}{7c+a}leqfrac{sqrt3}{8}$$
I tried C-S:
$$left(sum_{cyc}frac{1}{7a+b}right)^2leqsum_{cyc}frac{1}{(ka+mb+c)(7a+b)^2}sum_{cyc}(ka+mb+c)=$$
$$=sum_{cyc}frac{(k+m+1)(a+b+c)}{(ka+mb+c)(7a+b)^2}.$$
Thus, it remains to prove that
$$sum_{cyc}frac{k+m+1}{(ka+mb+c)(7a+b)^2}leqfrac{3}{64abc},$$
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
asked Mar 21 '17 at 20:05
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
|
show 2 more comments
$begingroup$
Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=abc$. Prove that:
$$frac{1}{7a+b}+frac{1}{7b+c}+frac{1}{7c+a}leqfrac{sqrt3}{8}$$
I tried C-S:
$$left(sum_{cyc}frac{1}{7a+b}right)^2leqsum_{cyc}frac{1}{(ka+mb+c)(7a+b)^2}sum_{cyc}(ka+mb+c)=$$
$$=sum_{cyc}frac{(k+m+1)(a+b+c)}{(ka+mb+c)(7a+b)^2}.$$
Thus, it remains to prove that
$$sum_{cyc}frac{k+m+1}{(ka+mb+c)(7a+b)^2}leqfrac{3}{64abc},$$
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
asked Mar 21 '17 at 20:05
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
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 $a=sqrt{3}x, b = sqrt{3}y, c = sqrt{3}z$ 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
yesterday
1
$begingroup$
@user574848 I tried. I think BW does not help here.
$endgroup$
– Michael Rozenberg
yesterday
|
show 2 more comments
$begingroup$
Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=abc$. Prove that:
$$frac{1}{7a+b}+frac{1}{7b+c}+frac{1}{7c+a}leqfrac{sqrt3}{8}$$
I tried C-S:
$$left(sum_{cyc}frac{1}{7a+b}right)^2leqsum_{cyc}frac{1}{(ka+mb+c)(7a+b)^2}sum_{cyc}(ka+mb+c)=$$
$$=sum_{cyc}frac{(k+m+1)(a+b+c)}{(ka+mb+c)(7a+b)^2}.$$
Thus, it remains to prove that
$$sum_{cyc}frac{k+m+1}{(ka+mb+c)(7a+b)^2}leqfrac{3}{64abc},$$
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
asked Mar 21 '17 at 20:05
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=abc$. Prove that:
$$frac{1}{7a+b}+frac{1}{7b+c}+frac{1}{7c+a}leqfrac{sqrt3}{8}$$
I tried C-S:
$$left(sum_{cyc}frac{1}{7a+b}right)^2leqsum_{cyc}frac{1}{(ka+mb+c)(7a+b)^2}sum_{cyc}(ka+mb+c)=$$
$$=sum_{cyc}frac{(k+m+1)(a+b+c)}{(ka+mb+c)(7a+b)^2}.$$
Thus, it remains to prove that
$$sum_{cyc}frac{k+m+1}{(ka+mb+c)(7a+b)^2}leqfrac{3}{64abc},$$
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
inequality contest-math
asked Mar 21 '17 at 20:05
Michael RozenbergMichael Rozenberg
109k1896201
asked Mar 21 '17 at 20:05
Michael RozenbergMichael Rozenberg
109k1896201
edited Mar 25 at 11:30
Michael Rozenberg
asked Mar 21 '17 at 20:05
Michael RozenbergMichael Rozenberg
109k1896201
asked Mar 21 '17 at 20:05
Michael RozenbergMichael Rozenberg
109k1896201
asked Mar 21 '17 at 20:05
Michael RozenbergMichael Rozenberg
109k1896201
109k1896201
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 $a=sqrt{3}x, b = sqrt{3}y, c = sqrt{3}z$ 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
yesterday
1
$begingroup$
@user574848 I tried. I think BW does not help here.
$endgroup$
– Michael Rozenberg
yesterday
|
show 2 more comments
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 $a=sqrt{3}x, b = sqrt{3}y, c = sqrt{3}z$ 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
yesterday
1
$begingroup$
@user574848 I tried. I think BW does not help here.
$endgroup$
– Michael Rozenberg
yesterday
4
4
$begingroup$
FOr those fans who hate abbreviations, C-S is Cauchy-Schwartz
$endgroup$
– Mark Fischler
Mar 21 '17 at 20:10
$begingroup$
FOr those fans who hate abbreviations, C-S is Cauchy-Schwartz
$endgroup$
– Mark Fischler
Mar 21 '17 at 20:10
5
5
$begingroup$
As a good start, take $a=sqrt{3}x, b = sqrt{3}y, c = sqrt{3}z$ 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
$begingroup$
As a good start, take $a=sqrt{3}x, b = sqrt{3}y, c = sqrt{3}z$ 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
1
$begingroup$
@W-t-P It's cyclic and not symmetric.
$endgroup$
– Michael Rozenberg
Mar 24 at 20:42
$begingroup$
@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
yesterday
$begingroup$
@MichaelRozenberg Maybe BW helps? Though it doesn't seem to be particularly elegant...
$endgroup$
– user574848
yesterday
1
1
$begingroup$
@user574848 I tried. I think BW does not help here.
$endgroup$
– Michael Rozenberg
yesterday
$begingroup$
@user574848 I tried. I think BW does not help here.
$endgroup$
– Michael Rozenberg
yesterday
|
show 2 more comments
0
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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 $a=sqrt{3}x, b = sqrt{3}y, c = sqrt{3}z$ 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
yesterday
1
$begingroup$
@user574848 I tried. I think BW does not help here.
$endgroup$
– Michael Rozenberg
yesterday