3 replies to this topic

[Forgive Yourself]

Cho $a,b,c$ dương thỏa mãn: $a+b+c=2$. Tìm GTNN: $P=\frac{a}{ab+2c}+\frac{b}{bc+2a}+\frac{c}{ca+2b}$

Edited by Forgive Yourself, 15-02-2013 - 18:15.

[ilovelife]

$$P = \sum \frac{1}{a+2.c/a} \ge \frac{(1+1+1)^2}{\sum a + 2(\frac ca + \frac bc + \frac ab)} \ge \frac{9}{\sum a + 2(\frac ca + \frac bc + \frac ab)}=\frac 9{2 + 2.3} = \frac 98$$
sorry, mình sai 1 chỗ

Edited by ilovelife, 15-02-2013 - 18:29.

[banhgaongonngon]

$$P = \sum \frac{1}{a+2.c/a} \ge \frac{(1+1+1)^2}{\sum a + 2(\frac ca + \frac bc + \frac ab)} \ge \frac{9}{\sum a + 2(\frac ca + \frac bc + \frac ab)}=\frac 9{2 + 2.3} = \frac 94$$
sorry, mình sai 1 chỗ

Mình thấy cách của bạn sai rồi, ở cái mẫu đó

[19kvh97]

Cho $a,b,c$ dương thỏa mãn: $a+b+c=2$. Tìm GTNN: $P=\frac{a}{ab+2c}+\frac{b}{bc+2a}+\frac{c}{ca+2b}$

$P=\sum \frac{a}{ab+(a+b+c)c}=\sum \frac{a}{(b+c)(a+c)}=\frac{1}{(a+b)(b+c)(c+a)}((a+b+c)^2-(ab+bc+ca))$
theo AM-GM
$\frac{1}{(a+b)(b+c)(c+a)}\geq \frac{27}{(2(a+b+c))^3}=\frac{27}{64}$
mà $ab+bc+ca\leq \frac{(a+b+c)^2}{3}=\frac{4}{3}$
nên $P\geq \frac{27}{64}(4-\frac{4}{3})=\frac{9}{8}$