\(T=\dfrac{a+b}{\sqrt{ab+c}}+\dfrac{b+c}{\sqrt{bc+a}}+\dfrac{c+a}{\sqrt{ca+b}}\)
\(\odot\) Ta có: \(\dfrac{a+b}{\sqrt{ab+c}}=\dfrac{a+b}{\sqrt{ab+c\left(a+b+c\right)}}=\dfrac{a+b}{\sqrt{\left(b+c\right)\left(a+c\right)}}\)
\(\odot\) Tương tự:
\(\dfrac{b+c}{\sqrt{bc+a}}=\dfrac{b+c}{\sqrt{\left(a+b\right)\left(a+c\right)}}\)
\(\dfrac{c+a}{\sqrt{ca+b}}=\dfrac{c+a}{\sqrt{\left(a+b\right)\left(b+c\right)}}\)
\(\odot\) Áp dụng bất đẳng thức AM - GM
\(\Rightarrow T=\dfrac{a+b}{\sqrt{\left(a+c\right)\left(b+c\right)}}+\dfrac{b+c}{\sqrt{\left(a+c\right)\left(b+a\right)}}+\dfrac{a+c}{\sqrt{\left(a+b\right)\left(b+c\right)}}\)
\(\ge3\sqrt[3]{\dfrac{a+b}{\sqrt{\left(a+c\right)\left(b+c\right)}}\times\dfrac{b+c}{\sqrt{\left(a+c\right)\left(b+a\right)}}\times\dfrac{a+c}{\sqrt{\left(a+b\right)\left(b+c\right)}}}\)
\(=3\)
\(\odot\) Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)
