\(\frac{a}{b+c}+1+\frac{b}{c+a}+1+\frac{c}{a+b}+1=\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)\)
\(\ge\left(a+b+c\right)\left(\frac{9}{b+c+c+a+a+b}\right)=\frac{\left(a+b+c\right)9}{2\left(a+b+c\right)}=\frac{9}{2}\)
\(\Rightarrow\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{9}{2}-3=\frac{3}{2}\)
\(VT=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
\(=\frac{a}{b+c}+1+\frac{b}{c+a}+1+\frac{c}{a+b}+1-3\)
\(=\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b}-3\)
\(=\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\)
\(=\frac{1}{2}[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)]\)\(\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)-3\)
C/m BĐT phụ \(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge9\text{(*) }\) với x, y, z dương
Áp dụng BĐT Cô-si ta có:
\(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge3\sqrt[3]{xyz}.3\sqrt[3]{\frac{1}{xyz}}=9\)
ÁP dụng BĐT (*) ta có:
\(VT=\frac{1}{2}\left[\left(x+y\right)+\left(y+z\right)+\left(z+x\right)\right]\)\(\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)-3\)
\(VT\ge\frac{1}{2}.9-3=\frac{3}{2}\left(đpcm\right)\)
\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{a^2}{a\left(b+c\right)}+\frac{b^2}{b\left(c+a\right)}+\frac{c^2}{c\left(a+b\right)}\)
>= \(\frac{\left(a+b+c\right)^2}{a\left(b+c\right)+b\left(c+a\right)+c\left(a+b\right)}=\frac{a^2+b^2+c^2+2ab+2ac+2bc}{2ab+2ac+2bc}\)(bđt cauchy schawatz dạng engel)(1)
mà \(a^2+b^2+c^2>=ab+ac+bc\)
\(\Rightarrow\frac{a^2+b^2+c^2+2ab+2ac+2bc}{2ab+2ac+2bc}>=\frac{ab+ac+bc+2ab+2ac+2bc}{2ab+2ac+2bc}\)
\(=\frac{3ab+3ac+3bc}{2ab+2ac+2bc}=\frac{3\left(ab+ac+bc\right)}{2\left(ab+ac+bc\right)}=\frac{3}{2}\)(2)
từ (1) và (2)\(\Rightarrow\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}>=\frac{3}{2}\)
dấu = xảy ra khi a=b=c
