Không mất tính tổng quát giả sử \(a\ge b\ge c>0\)
\(BĐT< =>\frac{a\left(b+c\right)\left(c+a\right)+b\left(a+b\right)\left(c+a\right)+c\left(a+b\right)\left(b+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge\frac{3}{2}\)
\(< =>\frac{ac^2+ba^2+cb^2+\left(a+b+c\right)\left(ab+bc+ca\right)}{\left(a+b+c\right)\left(ab+bc+ca\right)-abc}\ge\frac{3}{2}\)
\(< =>2\left[ac^2+ba^2+cb^2+\left(a+b+c\right)\left(ab+bc+ca\right)\right]\ge3\left[\left(a+b+c\right)\left(...\right)-abc\right]\)
\(< =>2\left(ac^2+a^2b+cb^2\right)\ge\left(a+b+c\right)\left(ab+bc+ca\right)-3abc\)
\(< =>ac^2+a^2b+cb^2\ge ca^2+ab^2+c^2b\)
\(< =>\left(c-b\right)\left(c-a\right)\left(a-b\right)\ge0\)(đúng)
Vậy ta có điều phải chứng minh
Ta có bất đẳng thức sau \(\left[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right]\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\ge9\)( cm = bunhia phân thức )
\(< =>1+\frac{a+b}{b+c}+\frac{a+b}{c+a}+1+\frac{b+c}{a+b}+\frac{b+c}{c+a}+1+\frac{c+a}{a+b}+\frac{c+a}{b+c}\ge9\)
\(< =>\frac{a}{a+b}+\frac{2a}{b+c}+\frac{a}{c+a}+\frac{b}{a+b}+\frac{b}{b+c}+\frac{2b}{c+a}+\frac{2c}{a+b}+\frac{c}{b+c}+\frac{c}{c+a}\ge6\)(*)
Đặt \(A=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\);\(B=\frac{a}{a+c}+\frac{b}{b+a}+\frac{c}{c+b}\);\(C=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
Khi đó bất đẳng thức (*) tương đương với \(A+B+2C\ge6\)
Do\(A+B=3\)\(=>2C\ge3=>C\ge\frac{3}{2}\)
Suy ra \(A+B+C\ge6-\frac{3}{2}=\frac{12-3}{2}=\frac{9}{2}\)(1)
Xét tổng :\(B+C=\frac{a}{a+c}+\frac{b}{b+a}+\frac{c}{c+b}+\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{a+b}{a+c}+\frac{c+a}{b+c}+\frac{b+c}{a+b}\ge3\)(AM-GM) (2)
Từ (1) và (2) ta được \(A\ge\frac{9}{2}-3=\frac{3}{2}\)
Done !
