+ Theo bđt Cauchy :
\(\sqrt{\left(a+b\right)\cdot\frac{2}{3}}\le\frac{a+b+\frac{2}{3}}{2}\) Dấu "=" \(\Leftrightarrow a+b=\frac{2}{3}\)
\(\sqrt{\left(b+c\right)\cdot\frac{2}{3}}\le\frac{b+c+\frac{2}{3}}{2}\) Dấu "=" \(\Leftrightarrow b+c=\frac{2}{3}\)
\(\sqrt{\left(c+a\right)\cdot\frac{2}{3}}\le\frac{c+a+\frac{2}{3}}{2}\) Dấu "=" \(\Leftrightarrow c+a=\frac{2}{3}\)
Do đó : \(\sqrt{\frac{2}{3}}Q\le\frac{2\left(a+b+c\right)+2}{2}=2\)
\(\Rightarrow Q\le\sqrt{6}\)
"=" \(\Leftrightarrow a=b=c=\frac{1}{3}\)