Cho a, b, c là các số thực dương. Chứng minh rằng:
$\frac{a}{3a+b+c}+\frac{b}{a+3b+c}+\frac{c}{a+b+3c}\geq \frac{24abc}{5(a+b)(b+c)(c+a)}$
$\frac{a}{3a+b+c}+\frac{b}{a+3b+c}+\frac{c}{a+b+3c}\geq \frac{24abc}{5(a+b)(b+c)(c+a)}$
Started By dinhkhanhly, 12-04-2016 - 15:31
#2
Posted 12-04-2016 - 20:59
leanhthu
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Binh nhất
BĐT $\Leftrightarrow \frac{1}{bc\cdot \left ( 3a+b+c \right )}+\frac{1}{ac\cdot \left ( a+3b+c \right )}+\frac{1}{ab\left ( a+b+3c \right )}\geq \frac{24}{5(a+b)\cdot \left ( b+c \right )\cdot \left ( c+a \right )}$
Áp dụng bất đẳng thức cauchy - schwartz, ta có:
$ \frac{1}{bc\cdot \left ( 3a+b+c \right )}+\frac{1}{ac\cdot \left ( a+3b+c \right )}+\frac{1}{ab\left ( a+b+3c \right )}\geqslant \frac{9}{(a+b)(b+c)(c+a)+7abc}$
ta có : $ \frac{9}{(a+b)(b+c)(c+a)+7abc}$$ \geq \frac{24}{5(a+b)(b+c)(c+a)}$$ \Leftrightarrow (a+b)(b+c)(c+a)\geq 8abc$
Dấu bằng <=> a=b=c
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