Question

Cho \(a>0,b>0,c>0\) thỏa mãn \(2\left(b^2+bc+c^2\right)=3\left(3-a^2\right)\)

Tìm GTNN của biểu thức \(T=a+b+c+\frac{2}{a}+\frac{2}{b}+\frac{2}{c}\)

Answer 1

\(9=3a^2+2b^2+2c^2+2bc=\left(a+b+c\right)^2+2a^2+b^2+c^2-2a\left(b+c\right)\)

\(\Rightarrow9\ge\left(a+b+c\right)^2+2a^2+\frac{1}{2}\left(b+c\right)^2-2a\left(b+c\right)\)

\(\Rightarrow9\ge\left(a+b+c\right)^2+2\sqrt{2a^2.\frac{1}{2}\left(b+c\right)^2}-2a\left(b+c\right)=\left(a+b+c\right)^2\)

\(\Rightarrow a+b+c\le3\)

\(T\ge a+b+c+\frac{18}{a+b+c}=a+b+c+\frac{9}{a+b+c}+\frac{9}{a+b+c}\)

\(T\ge2\sqrt{\frac{9\left(a+b+c\right)}{a+b+c}}+\frac{9}{3}=9\)

Dấu "=" xảy ra khi \(a=b=c=1\)