Áp dụng BĐT [TEX]Cauchy - Schwarz[/TEX] ta có:
[TEX]{\left[ {2\left( {a + b + c} \right) - abc} \right]^2} = {\left[ {a\left( {2 - bc} \right) + 2\left( {b + c} \right)} \right]^2} \leqslant \left( {{a^2} + 2} \right)\left[ {{{\left( {2 - bc} \right)}^2} + 2{{\left( {b + c} \right)}^2}} \right] = \left( {{a^2} + 2} \right)\left( {{b^2} + 2} \right)\left( {{c^2} + 2} \right)[/TEX]
Lại theo BĐT, ta có:
[TEX]\left( {{a^2} + 2} \right)\left( {{b^2} + 2} \right)\left( {{c^2} + 2} \right) = \dfrac{1}{6}\left( {3{a^2} + 6} \right)\left( {2{b^2} + 4} \right)\left( {{c^2} + 2} \right) \leqslant \dfrac{1}{6}{\left( {\dfrac{{3{a^2} + 2{b^2} + {c^2} + 12}}{3}} \right)^3} \leqslant 36[/TEX]
Tương đương: [TEX]{\left[ {2\left( {a + b + c} \right) - abc} \right]^2} \leqslant 36 \Rightarrow - 6 \leqslant 2\left( {a + b + c} \right) - abc \leqslant 6[/TEX]
Dấu "=" xảy ra khi và chỉ khi: [TEX]\left\{ \begin{array}{l}
a\left( {b + c} \right) = 2 - bc\\
3{a^2} + 6 = 2{b^2} + 4 = {c^2} + 2 = 6
\end{array} \right.[/TEX]
Tương đương: [TEX]\left[ \begin{array}{l}
a = 0;b = 1;c = 2\\
a = 0;b = - 1;c = - 2
\end{array} \right.[/TEX]
Vậy:
[TEX]\begin{array}{l}
Min\left[ {2\left( {a + b + c} \right) - abc} \right] = - 6 \Leftrightarrow a = 0;b = - 1;c = - 2\\
Max\left[ {2\left( {a + b + c} \right) - abc} \right] = 6 \Leftrightarrow a = 0;b = 1;c = 2
\end{array}[/TEX]