MÌnh nghĩ đề phải là tìm GTLN chứ
Ta có: \(\frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}=2\)
\(\Rightarrow\frac{1}{a+b+1}=\frac{b+c}{b+c+1}+\frac{c+a}{c+a+1}\ge2\sqrt{\frac{\left(b+c\right)\left(c+a\right)}{\left(b+c+1\right)\left(c+a+1\right)}}\)
Tương tự: \(\frac{1}{b+c+1}\ge2\sqrt{\frac{\left(a+b\right)\left(c+a\right)}{\left(a+b+1\right)\left(c+a+1\right)}}\)
\(\frac{1}{c+a+1}\ge2\sqrt{\frac{\left(a+b\right)\left(b+c\right)}{\left(a+b+1\right)\left(b+c+1\right)}}\)
Nhân lại ta có: \(\frac{1}{\left(a+b+1\right)\left(b+c+1\right)\left(c+a+1\right)}\ge\frac{8\left(a+b\right)\left(b+c\right)\left(c+a\right)}{\left(a+b+1\right)\left(b+c+1\right)\left(c+a+1\right)}\)
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\le\frac{1}{8}\)
Dấu = khi a=b=c=1/4
