$\frac{2a}{b+c}+\frac{2b}{a+c}+\frac{2c}{b+a}\geq 3+\frac{\left ( a-b \right )^{2}+\left ( b-c \right )^{2}+\left ( c-a \right )^{2}}{\left ( a+b+c \right )^{2}}$.
Edited by Ispectorgadget, 29-04-2012 - 17:03.
cmr $\sum \frac{2a}{b+c}\geq 3+\frac{\sum (a-b)^2}{(a+b+c)^2}$
Started By ironman, 29-04-2012 - 15:04
#2
Posted 29-04-2012 - 17:29
Giang1994
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C'est la vie
Chuẩn hóa $a+b+c=1$ ta có:
$\sum\frac{2a}{1-a}\geq 3 + \sum(a-b)^2$
$\sum\frac{2a}{1-a}\geq 3+ 2((a+b+c)^2-3(ab+bc+ca))$
$\sum\frac{2a}{1-a} \geq 5+ \sum3a(1-a)$
$ \sum \frac{2a}{1-a}-3a(1-a) \geq 5$
Chứng minh được
$\frac{2a}{1-a}-3a(1-a) \geq \frac{11a}{2}-\frac{1}{6}$
Suy ra dieu phai chứng minh
$\sum\frac{2a}{1-a}\geq 3 + \sum(a-b)^2$
$\sum\frac{2a}{1-a}\geq 3+ 2((a+b+c)^2-3(ab+bc+ca))$
$\sum\frac{2a}{1-a} \geq 5+ \sum3a(1-a)$
$ \sum \frac{2a}{1-a}-3a(1-a) \geq 5$
Chứng minh được
$\frac{2a}{1-a}-3a(1-a) \geq \frac{11a}{2}-\frac{1}{6}$
Suy ra dieu phai chứng minh
Edited by Giang1994, 29-04-2012 - 17:31.
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