\(VT=\frac{a}{a+\sqrt{a\left(a+b+c\right)+bc}}+\frac{b}{b+\sqrt{b\left(a+b+c\right)+ac}}+\frac{c}{c+\sqrt{c\left(a+b+c\right)+ab}}\)
\(VT=\frac{a}{a+\sqrt{\left(b+a\right)\left(a+c\right)}}+\frac{b}{b+\sqrt{\left(b+a\right)\left(c+b\right)}}+\frac{c}{c+\sqrt{\left(c+a\right)\left(b+c\right)}}\)
\(VT\le\frac{a}{a+\sqrt{\left(\sqrt{ab}+\sqrt{ac}\right)^2}}+\frac{b}{b+\sqrt{\left(\sqrt{ab}+\sqrt{bc}\right)^2}}+\frac{c}{c+\sqrt{\left(\sqrt{ac}+\sqrt{bc}\right)^2}}\)
\(VT\le\frac{a}{a+\sqrt{ab}+\sqrt{ac}}+\frac{b}{b+\sqrt{ab}+\sqrt{bc}}+\frac{c}{c+\sqrt{ac}+\sqrt{bc}}\)
\(VT\le\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}+\frac{\sqrt{b}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}+\frac{\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}=1\)
Dấu "=" xảy ra khi \(a=b=c=\frac{2}{3}\)
