Link: https://vn.answers.yahoo.com/question/index?qid=20100612215240AA1bp3C
Câu hỏi của Hạnh Tâm Nguyễn - Toán lớp 9 | Học trực tuyến
\(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}=\dfrac{a^2}{ab+ac}+\dfrac{b^2}{ab+bc}+\dfrac{c^2}{ac+bc}\)
Áp dụng BĐT Cauchy dạng Engel , ta có :
\(\dfrac{a^2}{ab+ac}+\dfrac{b^2}{ab+bc}+\dfrac{c^2}{ac+bc}\) ≥ \(\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ac\right)}\left(1\right)\)
Lại có : \(a^2+b^2+c^2\text{≥}ab+bc+ac\)
⇔ \(\left(a+b+c\right)^2\text{≥}3\left(ab+bc+ac\right)\left(2\right)\)
⇒ \(\dfrac{a^2}{ab+ac}+\dfrac{b^2}{ab+bc}+\dfrac{c^2}{ac+bc}\text{≥}\dfrac{3\left(ab+bc+ac\right)}{2\left(ab+bc+ac\right)}=\dfrac{3}{2}\)
\("="\text{⇔}a=b=c\)
CÁCH 1 :
Áp dụng bất đẳng thức cô - si cho 2 số không âm , ta có :
\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=\left(\dfrac{a}{b+c}+1\right)+\left(\dfrac{b}{c+a}+1\right)+\left(\dfrac{c}{a+b}+1\right)-3\)
\(=\left(a+b+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\right)-3\)
\(=\dfrac{1}{2}\left[\left(b+c\right)+\left(c+a\right)+\left(a+b\right)\right]\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\right)-3\)
\(=\dfrac{3}{2}+\dfrac{1}{2}\left(\dfrac{b+c}{c+a}+\dfrac{c+a}{b+c}\right)+\dfrac{1}{2}\left(\dfrac{c+a}{a+b}+\dfrac{a+b}{b+a}\right)+\dfrac{1}{2}\left(\dfrac{b+c}{a+b}+\dfrac{a+b}{b+c}\right)-3\)
\(\ge\dfrac{3}{2}+\sqrt{\dfrac{b+c}{c+a}.\dfrac{c+a}{b+c}}+\sqrt{\dfrac{c+a}{a+b}.\dfrac{a+b}{c+a}}+\sqrt{\dfrac{b+c}{a+b}.\dfrac{a+b}{b+c}}-3\)
\(=\dfrac{3}{2}\)
\(\Rightarrow\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge\dfrac{3}{2}\)
CÁCH 2 :
Áp dụng bất đẳng thức cô - si cho 2 số không âm , ta có :
\(\dfrac{a^2}{b+c}+\dfrac{b+c}{4}\ge2\sqrt{\dfrac{a^2}{b+c}.\dfrac{b+c}{4}}\)
\(\Rightarrow\dfrac{a^2}{b+c}+\dfrac{b+c}{4}\ge a\)
\(\Rightarrow\dfrac{a^2}{b+c}+a\ge2a-\dfrac{b+c}{4}\)
\(\Rightarrow\dfrac{a}{b+c}\left(a+b+c\right)\ge2a-\dfrac{b+c}{4}\)
Tương tự :
\(\dfrac{b}{c+a}\left(a+b+c\right)\ge2b-\dfrac{c+a}{4}\)
\(\dfrac{c}{a+b}\left(a+b+c\right)\ge2c-\dfrac{a+b}{4}\)
Ta có : \(\left(a+b+c\right)\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)\ge\dfrac{3}{2}\left(a+b+c\right)\)
\(\Rightarrow\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge\dfrac{3}{2}\left(đpcm\right)\)
