\(\Leftrightarrow\frac{a}{\sqrt{2b+2\sqrt{ca}}}+\frac{b}{\sqrt{2c+2\sqrt{ab}}}+\frac{c}{\sqrt{2a+2\sqrt{bc}}}\ge\frac{3}{2}\)
Ta có: \(abc=1\Rightarrow a+b+c\ge3\Rightarrow\left(a+b+c\right)^2\ge3\left(a+b+c\right)\)
\(VT\ge\frac{a}{\sqrt{a+2b+c}}+\frac{b}{\sqrt{a+b+2c}}+\frac{c}{\sqrt{2a+b+c}}\)
\(VT\ge\frac{4a}{2.2\sqrt{a+2b+c}}+\frac{4b}{2.2\sqrt{a+b+2c}}+\frac{4c}{2.2\sqrt{2a+b+c}}\)
\(VT\ge\frac{4a}{a+2b+c+4}+\frac{4b}{a+b+2c+4}+\frac{4c}{2a+b+c+4}=\frac{4a^2}{a^2+2ab+ac+4a}+\frac{4b^2}{ab+b^2+2bc+4b}+\frac{4c^2}{2ac+bc+c^2+4c}\)
\(VT\ge\frac{4\left(a+b+c\right)^2}{a^2+b^2+c^2+3\left(ab+bc+ca\right)+4\left(a+b+c\right)}\)
\(VT\ge\frac{4\left(a+b+c\right)^2}{\left(a+b+c\right)^2+ab+bc+ca+\frac{4}{3}.3\left(a+b+c\right)}\ge\frac{4\left(a+b+c\right)^2}{\left(a+b+c\right)^2+\frac{1}{3}\left(a+b+c\right)^2+\frac{4}{3}\left(a+b+c\right)^2}=\frac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
