- Với \(a,b,c>0\) và \(abc=1\), ta có:
\(\sqrt{\dfrac{a+b}{c+1}}+\sqrt{\dfrac{b+c}{a+1}}+\sqrt{\dfrac{c+a}{b+1}}\)
\(=\dfrac{a+b}{\sqrt{\left(c+1\right)\left(a+b\right)}}+\dfrac{b+c}{\sqrt{\left(a+1\right)\left(b+c\right)}}+\dfrac{c+a}{\sqrt{\left(b+1\right)\left(c+a\right)}}\)
\(\ge\dfrac{a+b}{\dfrac{\left(c+1\right)+\left(a+b\right)}{2}}+\dfrac{b+c}{\dfrac{\left(b+1\right)+\left(c+a\right)}{2}}+\dfrac{c+a}{\dfrac{\left(c+1\right)+\left(b+c\right)}{2}}\)
\(=\dfrac{2\left(a+b\right)}{a+b+c+1}+\dfrac{2\left(b+c\right)}{a+b+c+1}+\dfrac{2\left(c+a\right)}{a+b+c+1}\)
\(=\dfrac{4\left(a+b+c\right)}{a+b+c+1}\)
\(=4-\dfrac{4}{a+b+c+1}\ge4-\dfrac{4}{3\sqrt[3]{abc}+1}=4-\dfrac{4}{3\sqrt[3]{1}+1}=3\left(đpcm\right)\)
- Dấu "=" xảy ra khi \(a=b=c=1\)
