\(a,b,c>0\)
\(BĐT\Leftrightarrow\dfrac{a^2+2ab+b^2}{ab}+\dfrac{b^2+2bc+c^2}{bc}+\dfrac{c^2+2ca+a^2}{ca}\ge9+2\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)\)\(\Leftrightarrow\dfrac{a}{b}+2+\dfrac{b}{a}+\dfrac{b}{c}+2+\dfrac{c}{b}+\dfrac{c}{a}+2+\dfrac{a}{c}\ge9+2\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)\)
\(\Leftrightarrow a\left(\dfrac{1}{b}+\dfrac{1}{c}\right)+b\left(\dfrac{1}{c}+\dfrac{1}{a}\right)+c\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge3+2\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)\) (*)
Ta có, theo BĐT Caushy-Schwarz:
\(a\left(\dfrac{1}{b}+\dfrac{1}{c}\right)+b\left(\dfrac{1}{c}+\dfrac{1}{a}\right)+c\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge a.\dfrac{\left(1+1\right)^2}{b+c}+b.\dfrac{\left(1+1\right)^2}{c+a}+c.\dfrac{\left(1+1\right)^2}{a+b}=\dfrac{4a}{b+c}+\dfrac{4b}{c+a}+\dfrac{4c}{a+b}\)Mặt khác:
\(\dfrac{4a}{b+c}+\dfrac{4b}{c+a}+\dfrac{4c}{a+b}=2\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)+2\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)=2\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)+2\left(\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}+\dfrac{a+b+c}{a+b}-3\right)=2\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)+2\left(a+b+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\right)-6\)
\(\ge2\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)+2\left(a+b+c\right).\dfrac{\left(1+1+1\right)^2}{b+c+c+a+a+b}-6=2\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)+2\left(a+b+c\right).\dfrac{9}{2\left(a+b+c\right)}-6=2\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)+3\)\(\Rightarrow\)(*) đúng.
\(\Rightarrowđpcm\)
