\(\dfrac{1}{a+2b+c}+\dfrac{1}{b+2c+a}+\dfrac{1}{c+2a+b}< =\dfrac{1}{a+3b}+\dfrac{1}{b+3c}+\dfrac{1}{c+3a}\)
Có \(\dfrac{1}{b+2c+a}+\dfrac{1}{a+3b}< =\dfrac{4}{2a+4b+2c}=\dfrac{2}{a+2b+c}\)
Cm tương tự, ta có:
\(\dfrac{1}{c+2a+b}+\dfrac{1}{b+3c}< =\dfrac{2}{b+2c+a}\)\(\)
\(\dfrac{1}{a+2b+c}+\dfrac{1}{c+3a}< =\dfrac{2}{c+2a+b}\)
Cộng 2 vế của 3 BĐT với nhau, ta có:
\(\dfrac{1}{b+2c+a}+\dfrac{1}{a+3b}+\dfrac{1}{c+2a+b}+\dfrac{1}{b+3c}+\dfrac{1}{a+2b+c}+\dfrac{1}{c+3a}< =\dfrac{2}{a+2b+c}+\dfrac{2}{b+2c+a}+\dfrac{2}{c+2a+b}\)
\(\Leftrightarrow\left(\dfrac{1}{b+2c+a}+\dfrac{1}{c+2a+b}+\dfrac{1}{a+2b+c}\right)+\left(\dfrac{1}{a+3b}+\dfrac{1}{b+3c}+\dfrac{1}{c+3a}\right)< =\dfrac{2}{a+2b+c}+\dfrac{2}{b+2c+a}+\dfrac{2}{c+2a+b}\)
\(\Leftrightarrow\dfrac{-\left(c+2a+b\right)\cdot\left(a+2b+c\right)-\left(b+2c+a\right)\left(a+2b+c\right)-\left(b+2c+a\right)\left(c+2a+b\right)}{\left(b+2c+a\right)\cdot\left(c+2a+b\right)\cdot\left(a+2b+c\right)}+\dfrac{\left(b+3c\right)\left(c+3a\right)+\left(a+3b\right)\left(c+3a\right)+\left(a+3b\right)\left(b+3c\right)}{\left(a+3b\right)\left(b+3c\right)\left(c+3a\right)}\le0\)