Chủ đề này có 2 trả lời

[songviae]

Cho a,b,c>0; $a+b+c\leq ab+bc+ca$.CMR: $\frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+a}\leq 1$

[Dinh Xuan Hung]

Cho a,b,c>0; $a+b+c\leq ab+bc+ca$.CMR: $\frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+a}\leq 1$

BĐT cần chứng minh tương đương với:$\left ( 1-\frac{1}{1+a+b} \right )+\left ( 1-\frac{1}{1+b+c} \right )+\left ( 1-\frac{1}{1+c+a} \right )\geq 2$

$\Leftrightarrow \frac{a+b}{1+a+b}+\frac{b+c}{1+b+c}+\frac{c+a}{1+c+a}\geq 2$

$\Leftrightarrow \left ( \frac{a}{1+a+b}+\frac{b}{1+b+c}+\frac{c}{1+c+a} \right )+\left ( \frac{b}{1+a+b}+\frac{c}{1+b+c}+\frac{a}{1+a+c} \right )\geq 2(*)$

Mặt khác:

Áp dụng BĐT $C-S$ ta có:

$\frac{a}{1+a+b}+\frac{b}{1+b+c}+\frac{c}{1+c+a}=\frac{a^2}{a+a^2+ab}+\frac{b^2}{b+b^2+bc}+\frac{c^2}{c+ac+c^2}\geq \frac{(a+b+c)^2}{a^2+b^2+c^2+ab+bc+ac+a+b+c}\geq \frac{(a+b+c)^2}{a^2+b^2+c^2+2ab+2ac+2bc}=\frac{(a+b+c)^2}{(a+b+c)^2}=1$

$\frac{b}{1+a+b}+\frac{c}{1+b+c}+\frac{a}{1+a+c}\geq \frac{(a+b+c)^2}{b+ab+b^2+c+bc+c^2+a+a^2+ac}\geq \frac{(a+b+c)^2}{a^2+b^2+c^2+2(ab+bc+ac)}=\frac{(a+b+c)^2}{(a+b+c)^2}=1$

$\Rightarrow \left ( \frac{a}{1+a+b}+\frac{b}{1+b+c}+\frac{c}{1+c+a} \right )+\left ( \frac{b}{1+a+b}+\frac{c}{1+b+c}+\frac{a}{1+a+c} \right )\geq 2$

$\Rightarrow (*)$ luôn đúng 

Vậy $\frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+a}\leq 1$

Bài viết đã được chỉnh sửa nội dung bởi Dinh Xuan Hung: 15-08-2015 - 11:00

[hoanglong2k]

Cho a,b,c>0; $a+b+c\leq ab+bc+ca$.CMR: $\frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+a}\leq 1$

 Áp dụng Cauchy-Schwarz :

 $(1+a+b)(c^2+a+b)\geq (a+b+c)^2$

 $\Rightarrow \frac{1}{1+a+b}\leq \frac{c^2+a+b}{(a+b+c)^2}$

 $\Rightarrow \sum \frac{1}{1+a+b}\leq \frac{a^2+b^2+c^2+2(a+b+c)}{(a+b+c)^2}\leq 1$