If $a+b+c=abc$ then $sumlimits_{cyc}frac{1}{7a+b}leqfrac{sqrt3}{8}$
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.
Thank you!
inequality contest-math
asked Mar 21 '17 at 20:05
Michael Rozenberg
97k1589188
add a comment |
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.
Thank you!
inequality contest-math
asked Mar 21 '17 at 20:05
Michael Rozenberg
97k1589188
4
FOr those fans who hate abbreviations, C-S is Cauchy-Schwartz
– Mark Fischler
Mar 21 '17 at 20:10
4
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.
– Mark Fischler
Mar 21 '17 at 22:57
add a comment |
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.
Thank you!
inequality contest-math
asked Mar 21 '17 at 20:05
Michael Rozenberg
97k1589188
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.
Thank you!
inequality contest-math
inequality contest-math
asked Mar 21 '17 at 20:05
Michael Rozenberg
97k1589188
asked Mar 21 '17 at 20:05
Michael Rozenberg
97k1589188
asked Mar 21 '17 at 20:05
Michael Rozenberg
97k1589188
asked Mar 21 '17 at 20:05
Michael Rozenberg
97k1589188
asked Mar 21 '17 at 20:05
Michael Rozenberg
97k1589188
97k1589188
4
FOr those fans who hate abbreviations, C-S is Cauchy-Schwartz
– Mark Fischler
Mar 21 '17 at 20:10
4
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.
– Mark Fischler
Mar 21 '17 at 22:57
add a comment |
4
FOr those fans who hate abbreviations, C-S is Cauchy-Schwartz
– Mark Fischler
Mar 21 '17 at 20:10
4
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.
– Mark Fischler
Mar 21 '17 at 22:57
4
4
FOr those fans who hate abbreviations, C-S is Cauchy-Schwartz
– Mark Fischler
Mar 21 '17 at 20:10
FOr those fans who hate abbreviations, C-S is Cauchy-Schwartz
– Mark Fischler
Mar 21 '17 at 20:10
4
4
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.
– Mark Fischler
Mar 21 '17 at 22:57
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.
– Mark Fischler
Mar 21 '17 at 22:57
add a comment |
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4
FOr those fans who hate abbreviations, C-S is Cauchy-Schwartz
– Mark Fischler
Mar 21 '17 at 20:10
4
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.
– Mark Fischler
Mar 21 '17 at 22:57