Ta có:\(\dfrac{1}{a+b+1}+\dfrac{1}{b+c+1}+\dfrac{1}{a+c+1}=2\)
\(\Rightarrow\dfrac{1}{a+b+1}=\left(1-\dfrac{1}{b+c+1}\right)+\left(1-\dfrac{1}{a+c+1}\right)\)
\(\Rightarrow\dfrac{1}{a+b+1}=\dfrac{b+c}{b+c+1}+\dfrac{a+c}{a+c+1}\ge2\sqrt{\dfrac{\left(b+c\right)\left(a+c\right)}{\left(b+c+1\right)\left(a+c+1\right)}}\)Chứng minh tương tự :\(\dfrac{1}{b+c+1}\ge2\sqrt{\dfrac{\left(a+b\right)\left(a+c\right)}{\left(a+b+1\right)\left(a+c+1\right)}}\)
\(\dfrac{1}{a+c+1}\ge2\sqrt{\dfrac{\left(a+b\right)\left(b+c\right)}{\left(a+b+1\right)\left(b+c+1\right)}}\)
Nhân các bất đẳng thức trên lại với nhau về theo vế ,ta được:
\(\dfrac{1}{\left(a+b+1\right)\left(b+c+1\right)\left(a+c+1\right)}\ge\dfrac{8\left(a+b\right)\left(b+c\right)\left(c+a\right)}{\left(a+b+1\right)\left(b+c+1\right)\left(a+c+1\right)}\)
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\le\dfrac{1}{8}\)
Dấu "=" xảy ra khi:\(a=b=c=\dfrac{1}{4}\)
Vậy giá trị lớn nhất của (a+b)(b+c)(c+a) là \(\dfrac{1}{8}\) khi \(a=b=c=\dfrac{1}{4}\)
