Cho các số thực dương $a,b,c$, chứng minh rằng \[\dfrac{1}{13a+11b}+\dfrac{1}{13b+11c}+\dfrac{1}{13c+11a}\geq \dfrac{1}{7a+4b+13c}+\dfrac{1}{7b+4c+13a}+\dfrac{1}{7c+4a+13b}\]
$\sum \dfrac{1}{13a+11b}\geq \sum \dfrac{1}{7a+4b+13c}$
Started By Ankh, 11-04-2016 - 23:25
#1
Posted 11-04-2016 - 23:25
#2
Posted 23-06-2016 - 21:25
Áp dụng bđt cauchy-schwarz:
$\frac{(-8)^2}{-8(11a+11b)} + \frac{52^2}{52(13b+11c)} + \frac{103^2}{103(13c+11a)} \geq \frac{147^2}{147(7a+4b+13c)}$
Tương tự vs mấy cái còn lại, cộng vào ta đc:
$\sum\frac{147}{13a+11b} \geq \sum \frac{147}{7a+4b+13c}$
từ đó suy ra đpcm
Edited by nguyengoldz, 23-06-2016 - 21:27.
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