Chủ đề này có 2 trả lời

[zack]

cho a,b,c>0 thỏa ab+bc+ac=3.CMR

1.$\frac{1}{1+a(b+c)}+\frac{1}{1+b(a+c)}+\frac{1}{1+c(b+a)}\leq \frac{1}{abc}$

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

Bài viết đã được chỉnh sửa nội dung bởi zack: 25-04-2014 - 13:09

[raquaza]

cho a,b,c>0 thỏa ab+bc+ac=3.CMR

$\frac{1}{1+a(b+c)}+\frac{1}{1+b(a+c)}+\frac{1}{1+c(b+a)}\leq \frac{1}{abc}$

$\sum \frac{1}{1+ab+ac}=\sum \frac{2+bc}{(1+ab+ac)(bc+1+1)}\leq \sum \frac{2+bc}{(\sqrt{ab}+\sqrt{ac}+\sqrt{bc})^{2}}=\frac{6+ab+bc+ac}{(\sqrt{ab}+\sqrt{ac}+\sqrt{bc})^{2}}\leq \frac{9}{9\sqrt[3]{a^{2}b^{2}c^{2}}}\leq \frac{1}{abc}$

(vi ab+bc+ac=3$\geq 3\sqrt[3]{a^{2}b^{2}c^{2}}$ suy ra$abc\leq 1$)

[Kaito Kuroba]

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

2.

từ $$ab+bc+ca=3\Rightarrow abc\leq 1$$

từ đây ta có: $$\sum \frac{1}{1+a^2(b+c)}\leq \sum\frac{1}{a}. \frac{1}{bc+ab+ac}= \left ( \sum \frac{1}{a} \right ).\frac{1}{ab+bc+ca}=\frac{1}{abc}. "="\Leftrightarrow a=b=c=1$$