Giả sử \(a\ge b\ge c\)
Ta có:\(\frac{a+b}{ab+c^2}+\frac{b+c}{bc+a^2}+\frac{c+a}{ca+b^2}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
\(\Leftrightarrow\frac{ac+bc-ab-c^2}{c\left(ab+c^2\right)}+\frac{ab+ac-bc-a^2}{\left(bc+a^2\right)a}+\frac{cb+ab-ca-b^2}{b\left(ca+b^2\right)}\ge0\)
\(\Leftrightarrow\frac{\left(a-c\right)\left(c-b\right)}{c\left(ab+c^2\right)}+\frac{\left(b-a\right)\left(a-c\right)}{\left(bc+a^2\right)a}+\frac{\left(c-b\right)\left(b-a\right)}{b\left(ca+b^2\right)}\le0\)
Ta có:\(\left(c-b\right)\left(b-a\right)\ge0;\left(b-a\right)\left(a-c\right)\le0;\left(a-c\right)\left(c-b\right)\le0\)
\(\Rightarrow\frac{\left(c-b\right)\left(c-a\right)}{b\left(ca+b^2\right)}\le\frac{\left(c-b\right)\left(c-a\right)}{c\left(ab+c^2\right)}\)
\(\Rightarrow LHS\le\frac{\left(a-c\right)\left(c-b\right)}{c\left(ab+c^2\right)}+\frac{\left(c-b\right)\left(b-a\right)}{c\left(ab+c^2\right)}+\frac{\left(b-a\right)\left(a-c\right)}{\left(bc+a^2\right)a}\)
\(=\frac{-\left(c-b\right)^2}{c\left(ab+c^2\right)}+\frac{\left(b-a\right)\left(a-c\right)}{\left(bc+a^2\right)c}\le0\)
\(\Rightarrowđpcm\)
Trả lời :
Bn Thu Trang đừng bình luận linh tinh nhé.
- Hok tốt !
^_^
Đặt VT là A, VP là B
Xét hiệu A-B=\(\left(\frac{a+b}{ab+c^2}-\frac{1}{a}\right)+\left(\frac{b+c}{bc+a^2}-\frac{1}{b}\right)+\left(\frac{c+a}{ac+b^2}-\frac{1}{c}\right)\)
\(=\frac{\left(a-c\right)\left(c-b\right)}{\left(ab+c^2\right)c}+\frac{\left(b-a\right)\left(a-c\right)}{\left(bc+a^2\right)a}+\frac{\left(c-b\right)\left(b-a\right)}{\left(ca+b^2\right)b}\)
Do vai trò của phương trình là bình đẳng nên ta giả sử \(a\ge b\ge c\ge0\)
Khi đó \(\left(b-a\right)\left(a-c\right)\le0;\left(c-b\right)\left(b-a\right)\le0\)và \(c^3\le b^3\)
=> abc+c3 =< abc+b3
=> \(\frac{\left(c-b\right)\left(b-a\right)}{\left(ca+b^2\right)b}\le\frac{\left(c-b\right)\left(b-a\right)}{\left(ab+c^2\right)c}\)
Vậy \(A-B\le\frac{\left(a-c\right)\left(c-b\right)}{\left(ab+c^2\right)c}+\frac{\left(c-b\right)\left(b-a\right)}{\left(ab+c^2\right)c}+\frac{\left(b-a\right)\left(a-c\right)}{\left(bc+a^2\right)a}\)
\(=\frac{\left(a-c\right)\left(c-b\right)+\left(c-b\right)\left(b-a\right)}{\left(ab+c^2\right)c}+\frac{\left(b-a\right)\left(a-c\right)}{\left(bc+a^2\right)a}\)
\(=\frac{-\left(c-b\right)^2}{\left(ab+a^2\right)a}+\frac{\left(b-a\right)\left(a-c\right)}{\left(bc+a^2\right)a}\le0\)
Vì -(c-b)2 =< 0; (b-a)(a-c) =< 0
Vậy phương trình được chứng minh
Đào Nhật Quỳnh Cách cậu khác gì cách tớ đâu,được cái phần đầu tớ dùng dấu ngu ngu :)) Tớ làm tiếp cách khác nhé !
Theo BĐT Cauchy - Schwarz ta dễ có:\(\frac{a+b}{ab+c^2}=\frac{\left(a+b\right)^2}{\left(ab+c^2\right)\left(a+b\right)}=\frac{\left(a+b\right)^2}{a\left(b^2+c^2\right)+b\left(a^2+c^2\right)}\le\frac{a^2}{b\left(a^2+c^2\right)}+\frac{b^2}{a\left(b^2+c^2\right)}\)
\(\Rightarrow LHS\le\Sigma\left[\frac{a^2}{b\left(a^2+c^2\right)}+\frac{b^2}{a\left(b^2+c^2\right)}\right]=\Sigma\left[\frac{a^2}{b\left(a^2+c^2\right)}+\frac{c^2}{b\left(a^2+c^2\right)}\right]=\Sigma\frac{1}{b}\)
Vậy ta có đpcm
