Edited by Phạm Quang Toàn, 14-01-2013 - 21:35.
CMR: $\sum \frac{a+bc}{b+c}\geq 2$
#1
Posted 14-01-2013 - 20:16
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#2
Posted 14-01-2013 - 20:25
Ko hiểu sao mình thấy câu này rất na ná câu bđt số TTT2 kì này nên xin phép lock để bảo đám tính "an toàn" nhéCho $a,b,c>0$ và $a+b+c=1$. CMR: $\frac{a+bc}{b+c}+\frac{b+ca}{c+a}+\frac{c+ab}{a+b}\geq 2$
#3
Posted 14-01-2013 - 20:38
Bài TTT2 là $$\sum \frac{ab+c}{c+1} \le 1$$
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#4
Posted 14-01-2013 - 20:44
$$\frac{(b+c)(c+a)}{a+b}+\frac{(a+c)(a+b)}{b+c}+\frac{(a+b)(b+c)}{a+c}\geq 2$$
Đặt $a+b=z,b+c=x,c+a=y$ thì $x+y+z=2$ và BĐT cần c/minh trở thành
$$x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2}\geq xyz(x+y+z)$$
Đúng theo AM-GM
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